Explanation:
Okay, so I need to find the product of the y-coordinates of all the distinct solutions (x, y) for the two equations y = x² - 8 and y² = -5x + 44. Hmm, let me think about how to approach this.

First, since both equations involve y and x, maybe I can substitute one into the other. The first equation gives me y in terms of x, so maybe I can plug that expression for y into the second equation. Let me try that.

So, if y = x² - 8, then substituting into the second equation y² = -5x + 44 gives me (x² - 8)² = -5x + 44. Let me expand the left side.

Expanding (x² - 8)²: that's x⁴ - 16x² + 64. So the equation becomes x⁴ - 16x² + 64 = -5x + 44.

Now, let me bring all terms to one side to set the equation to zero. Subtract -5x and 44 from both sides:

x⁴ - 16x² + 64 + 5x - 44 = 0.

Simplify the constants: 64 - 44 = 20, so the equation becomes:

x⁴ - 16x² + 5x + 20 = 0.

Hmm, so now I have a quartic equation: x⁴ - 16x² + 5x + 20 = 0. Solving quartic equations can be tricky. Maybe I can factor this equation? Let me check for rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term (20) divided by factors of the leading coefficient (1), so possible roots are ±1, ±2, ±4, ±5, ±10, ±20.

Let me test x = 1: 1 - 16 + 5 + 20 = 1 -16 is -15, plus 5 is -10, plus 20 is 10. Not zero.

x = -1: (-1)^4 -16(-1)^2 +5(-1) +20 = 1 -16(1) -5 +20 = 1 -16 -5 +20 = 0. Oh! x = -1 is a root. Great! So (x + 1) is a factor.

Now, let me perform polynomial division or use synthetic division to factor out (x + 1) from the quartic.

Using synthetic division for root -1:

Coefficients of the quartic: 1 (x⁴), 0 (x³), -16 (x²), 5 (x), 20 (constant).

Set up synthetic division:

-1 | 1   0   -16   5   20
    |     -1   1   -15   -20
    ---------------------------
      1  -1   -15   -10   0

So the quartic factors into (x + 1)(x³ - x² -15x -10). Now, I need to factor the cubic equation x³ - x² -15x -10. Let me check for rational roots again. Possible roots are ±1, ±2, ±5, ±10.

Testing x = -1: (-1)^3 - (-1)^2 -15(-1) -10 = -1 -1 +15 -10 = 3 ≠ 0.

x = 2: 8 - 4 - 30 -10 = 8 -4 is 4, 4 -30 is -26, -26 -10 is -36 ≠ 0.

x = 5: 125 -25 -75 -10 = 125 -25 is 100, 100 -75 is 25, 25 -10 is 15 ≠ 0.

x = -2: (-8) - 4 - (-30) -10 = -8 -4 +30 -10 = 8 ≠ 0.

x = 10: That's going to be huge, probably not zero. Let me check x = - something else?

Wait, maybe x = 5 didn't work. How about x = - something?

Wait, maybe I missed something. Let me try x = -2 again:

(-2)^3 - (-2)^2 -15*(-2) -10 = -8 -4 +30 -10 = (-12) + 20 = 8. Not zero.

x = 5: already tried. How about x = - something else?

Wait, perhaps x = 5? No. Maybe x = 2? Tried. How about x = - something?

Wait, perhaps there is no rational root here? Maybe I need to use the rational root theorem again. Maybe I made an error in calculation?

Wait, let me check x = -2 again.

x³ -x² -15x -10 at x = -2:

(-8) - (4) -15*(-2) -10 = -8 -4 +30 -10 = (-8 -4 -10) +30 = (-22) +30 = 8. Correct. Not zero.

Hmm. Maybe this cubic factors into (x^3 - x² -15x -10). Let me try grouping terms?

x³ - x² -15x -10. Let me try grouping first two terms and last two terms:

(x³ - x²) + (-15x -10) = x²(x -1) -5(3x + 2). Doesn't seem helpful. Maybe different grouping?

Alternatively, perhaps there is a typo in the synthetic division? Let me check again.

Original quartic: x⁴ - 16x² + 5x + 20.

Divided by (x + 1):

Using synthetic division:

Coefficients: 1 (x^4), 0 (x^3), -16 (x^2), 5 (x), 20 (constant)

Bring down 1.

Multiply by -1: 1*(-1) = -1. Add to next coefficient: 0 + (-1) = -1.

Multiply by -1: -1*(-1) = 1. Add to next coefficient: -16 + 1 = -15.

Multiply by -1: -15*(-1) = 15. Add to next coefficient: 5 + 15 = 20.

Multiply by -1: 20*(-1) = -20. Add to last coefficient: 20 + (-20) = 0. Correct. So the cubic is x³ - x² -15x -10. So that's correct.

Since rational roots aren't working, maybe I need to use the method for solving cubics? Or maybe there's another factor?

Alternatively, maybe I made an error earlier in substituting? Let me check that again.

Original equations: y = x² -8 and y² = -5x +44.

Substitute y into second equation: (x² -8)^2 = -5x +44. So expanding left side: x^4 - 16x² + 64. Then moving all to left: x^4 -16x² +64 +5x -44 = x^4 -16x² +5x +20. Correct. So quartic equation is correct. So the cubic is correct.

Alternatively, maybe there is a typo? Wait, maybe I need to try another approach? Let me think.

Alternatively, maybe instead of substituting y from the first equation into the second, I can solve for x in terms of y from the second equation and substitute into the first? Let me try that.

From the second equation: y² = -5x + 44 => -5x = y² -44 => x = (44 - y²)/5.

Then substitute this x into the first equation y = x² -8:

So y = [(44 - y²)/5]^2 - 8.

Let me compute that:

First, square (44 - y²)/5: [ (44 - y²)² ] / 25. Then subtract 8:

y = [ (44 - y²)² / 25 ] - 8.

Multiply both sides by 25 to eliminate denominator:

25y = (44 - y²)² - 200.

Then expand (44 - y²)²: that is y^4 - 88y² + 44² = y^4 - 88y² + 1936.

Therefore, equation becomes:

25y = y^4 -88y² +1936 -200 => 25y = y^4 -88y² + 1736.

Bring all terms to left side:

0 = y^4 -88y² +1736 -25y => y^4 -88y² -25y +1736 = 0.

So quartic in y: y^4 -88y² -25y +1736 = 0. Hmm, not sure if this is easier. Maybe?

But earlier, we had a quartic in x: x^4 -16x² +5x +20 = 0, and we had found one root x=-1. Maybe trying to find solutions in y?

Alternatively, maybe using Vieta's formula? Wait, but if I can find the roots of the quartic in x, then for each x, I can find y from y = x² -8, then multiply the y's. However, since some solutions might not be distinct? Wait, the problem says "distinct solutions (x, y)", so even if two different x's give same y, they are considered distinct if their x's are different.

But since we need the product of the y-coordinates of all distinct solutions. So if I can find all solutions (x, y), then take their y's and multiply them.

But maybe there's a smarter way than solving quartic? Let me think.

Alternatively, if we have two equations:

y = x² -8

and

y² = -5x +44.

Let me consider that if I can find the solutions, and then compute the product of their y's.

Alternatively, maybe I can use Vieta's theorem on the quartic? Wait, but Vieta gives the product of all roots, but here we have a quartic in x. But the question is about the product of y's. However, maybe we can relate the roots in x to the y's?

Each solution (x, y) corresponds to an x that is a root of the quartic. For each such x, y = x² -8. Therefore, the product of all y's would be the product over each root x_i of (x_i² -8).

So if we can find the quartic equation's roots, then compute the product of (x_i² -8) for each root x_i. However, computing that directly might be complicated. Let me think if there's a way to relate this product to the coefficients of the quartic.

Alternatively, note that if we have a quartic equation x⁴ + a x³ + b x² + c x + d = 0, and we need to find the product of (x_i² -k) for all roots x_i. But that might be complex, but maybe we can express it in terms of the roots?

Alternatively, maybe use the fact that if we have the quartic equation, and we know that for each root x, y = x² -8. So maybe consider the product of (x² -8) over all roots x.

But the product of (x² -8) for all roots x is equal to the product over all roots of (x - sqrt(8))(x + sqrt(8)) = product (x - sqrt(8)) * product (x + sqrt(8)) = [product (x - sqrt(8))] [product (x + sqrt(8))].

But product (x - sqrt(8)) is equal to the value of the quartic evaluated at x = sqrt(8), with sign? Wait, actually, if we have a quartic equation f(x) = x⁴ -16x² +5x +20. Then f(sqrt(8)) would be?

But if the quartic factors as (x - x1)(x -x2)(x -x3)(x -x4), then product (x - sqrt(8)) = f(sqrt(8)) if the quartic is monic. Wait, actually, f(sqrt(8)) = (sqrt(8) -x1)(sqrt(8)-x2)(sqrt(8)-x3)(sqrt(8)-x4). Similarly, f(-sqrt(8)) = (-sqrt(8) -x1)(-sqrt(8)-x2)... So then the product of (x_i² - 8) is product (x_i - sqrt(8))(x_i + sqrt(8)) = product (x_i - sqrt(8)) * product (x_i + sqrt(8)) = f(sqrt(8)) * f(-sqrt(8)).

Therefore, if I compute f(sqrt(8)) * f(-sqrt(8)), that would give the product of all (x_i² -8), which is the product of all y's. So that might be a way to compute the desired product without finding all roots. Let me try that.

First, let me recall that f(x) = x⁴ -16x² +5x +20.

Compute f(sqrt(8)) and f(-sqrt(8)).

First, compute f(sqrt(8)):

Let me compute term by term.

x⁴: (sqrt(8))⁴ = (8)² = 64.

-16x²: -16*(sqrt(8))² = -16*8 = -128.

+5x: 5*sqrt(8)

+20

So total f(sqrt(8)) = 64 - 128 + 5*sqrt(8) +20 = (64 - 128 +20) + 5*sqrt(8) = (-44) + 5*sqrt(8)

Similarly, f(-sqrt(8)):

x⁴ is same as before, 64.

-16x² is same as -128.

+5x is 5*(-sqrt(8)) = -5*sqrt(8)

+20.

So total f(-sqrt(8)) = 64 -128 -5*sqrt(8) +20 = (-44) -5*sqrt(8)

Therefore, the product f(sqrt(8)) * f(-sqrt(8)) is [(-44) +5√8][(-44) -5√8] which is (-44)^2 - (5√8)^2.

Compute that:

(-44)^2 is 1936.

(5√8)^2 is 25*(8) = 200.

Therefore, the product is 1936 - 200 = 1736.

Wait, so according to this, the product of all y's is 1736? But hold on, is that correct?

Wait, but this is under the assumption that all roots are distinct? Wait, but the original problem mentions "distinct solutions (x, y)", so if there are multiple roots, maybe some y's are repeated? But Vieta's formula counts all roots, including multiplicities, so if there are multiple roots, the product would still be 1736? However, the problem says "distinct solutions", so I need to check if there are any multiple roots, which would result in duplicate solutions (x,y). However, since the quartic equation might have multiple roots, but since we have f(sqrt(8)) * f(-sqrt(8)) is 1736, but maybe some of the roots are repeated?

But actually, the problem says "distinct solutions", so even if the quartic has multiple roots, we need to take the product of y-coordinates of distinct solutions. So, if the quartic has a multiple root, that would correspond to a solution (x,y) that is repeated, but since we need distinct solutions, we have to consider only unique (x,y) pairs. Therefore, if there are multiple roots, they would give same (x,y) if they are same x. Wait, but for each root x, even if repeated, the corresponding y is x² -8, which would be same if x is same. Therefore, if there are multiple roots, say, x1 with multiplicity k, then the solution (x1, y) would be counted once, but the product would be over all distinct y's.

But here, my calculation gives the product over all roots (counting multiplicities) of (x_i² -8). Therefore, if there are multiple roots, the product would include the y corresponding to those multiple roots multiple times. Therefore, this approach would give the product over all roots, including multiplicities. However, the problem asks for the product of y-coordinates of distinct solutions. So if there are multiple roots, that would mean multiple x's with same y? Wait, no. If x is repeated, then y = x² -8 would be same for same x. Therefore, if x is a multiple root, then the same (x,y) is counted multiple times, but in the problem, we need to consider distinct solutions. Therefore, the product of distinct y-coordinates would be different. So this method would overcount?

Therefore, maybe this approach is incorrect?

Wait, let me think again. Let me denote that each solution (x,y) corresponds to a unique x. If there are multiple x's, even if same y, they are considered distinct solutions. Wait, for example, suppose x1 and x2 are different, but y1 = y2. Then (x1, y1) and (x2, y2) are distinct solutions. Therefore, even if same y, different x's give distinct solutions. Therefore, the number of solutions is equal to the number of roots of the quartic equation. However, if there are multiple roots (i.e., same x with multiplicity), then that would correspond to same solution (x,y) repeated. Therefore, the problem says "distinct solutions", so if quartic has multiple roots, the corresponding solutions would be same. Therefore, the product of the y-coordinates of distinct solutions would be the product over distinct y's, which may involve grouping same y's.

But how can I know which y's are repeated?

Alternatively, maybe the quartic is separable, meaning all roots are distinct? If that is the case, then the product of y's would be 1736. However, we need to check if there are multiple roots. To check that, perhaps compute the greatest common divisor of the quartic and its derivative? Let me try that.

Original quartic: f(x) = x⁴ -16x² +5x +20.

Compute its derivative: f’(x) = 4x³ -32x +5.

Now, check if there is a common factor between f(x) and f’(x). For that, let me use the Euclidean algorithm.

First, divide f(x) by f’(x). But since f(x) is degree 4 and f’(x) is degree 3, we can perform polynomial division.

Alternatively, find if f’(x) has any roots that are also roots of f(x). Let's check if x = -1 is a root of f’(x). Wait, x = -1 is a root of f(x). Let me compute f’(-1):

4*(-1)^3 -32*(-1) +5 = -4 +32 +5 = 33 ≠ 0. Therefore, x = -1 is not a root of f’(x). Therefore, unless there is another common root, which would have to be a root of both f and f’.

Alternatively, since we already factored f(x) as (x +1)(x³ -x² -15x -10). Let me check if the cubic x³ -x² -15x -10 has any multiple roots. Let me compute the derivative of the cubic:

g(x) = x³ -x² -15x -10, so g’(x) = 3x² - 2x -15.

Now, check if any roots of g(x) are also roots of g’(x). For that, solve g(x) = 0 and g’(x) = 0. If there's a common root, then it's a multiple root.

Let me suppose that α is a common root. Then:

α³ - α² -15α -10 = 0,

and

3α² - 2α -15 = 0.

Let me solve the second equation for α²:

3α² = 2α +15 => α² = (2α +15)/3.

Substitute into first equation:

α³ - α² -15α -10 = 0.

Let me express α³ as α * α². Since α² = (2α +15)/3, then:

α³ = α*(2α +15)/3 = (2α² +15α)/3. Substitute α² again:

= [2*(2α +15)/3 +15α]/3 = [ (4α +30)/3 +15α ] /3 = [ (4α +30 +45α)/3 ] /3 = [49α +30]/9.

Therefore, α³ = (49α +30)/9.

Now substitute into original equation:

α³ - α² -15α -10 = [ (49α +30)/9 ] - [ (2α +15)/3 ] -15α -10 = 0.

Multiply all terms by 9 to eliminate denominators:

49α +30 - 3*(2α +15) - 135α -90 = 0.

Compute term by term:

First term: 49α +30

Second term: -3*(2α +15) = -6α -45

Third term: -135α

Fourth term: -90

Combine:

49α +30 -6α -45 -135α -90 = (49α -6α -135α) + (30 -45 -90) = (-92α) + (-105) = 0.

Therefore, -92α -105 = 0 => α = -105/92 ≈ -1.141...

But this is a real number. Wait, does this mean that there is a common root? Wait, but unless this α satisfies both equations. Wait, but we derived it under the assumption that α is a common root. However, we need to check if this α actually satisfies both equations. Let me check if plugging α = -105/92 into the second equation, which was 3α² -2α -15 = 0. Let me compute:

First, compute 3α² -2α -15:

Let α = -105/92.

Compute α²: (105^2)/(92^2) = 11025 / 8464.

Multiply by 3: 33075 / 8464.

Compute -2α: -2*(-105/92) = 210/92 = 105/46.

Therefore, 3α² -2α -15 = 33075/8464 + 105/46 -15.

Convert all terms to have denominator 8464:

33075/8464 + (105/46)*(184/184) -15*(8464/8464)

Wait, 46*184 = 46*(180 +4) = 46*180 + 46*4 = 8280 + 184 = 8464. Yes.

So:

33075/8464 + (105*184)/8464 - (15*8464)/8464

Calculate numerator:

33075 + 105*184 -15*8464.

First compute 105*184:

105*184 = (100 +5)*184 = 100*184 + 5*184 = 18400 + 920 = 19320.

Then compute 15*8464: 15*8000 = 120000; 15*464= 6960 → total 120000 + 6960 = 126960.

Thus numerator: 33075 +19320 -126960 = (33075 +19320) -126960 = 52395 -126960 = -74565.

Therefore, 3α² -2α -15 = -74565 / 8464 ≈ -8.817 ≠ 0. Therefore, contradiction. So there is no common root between g(x) and g’(x). Therefore, the cubic has no multiple roots. Therefore, the original quartic equation has simple roots? Wait, since we factored it as (x +1)*(cubic), and the cubic has no multiple roots. Also, x = -1 is not a root of the derivative. Therefore, all roots are distinct. Hence, the quartic equation has four distinct real roots? Or maybe some complex roots?

Wait, but if the quartic equation has four distinct roots, then the original system has four distinct solutions (x,y). Therefore, the product of y-coordinates would be 1736. But wait, in my earlier calculation, I found that the product is 1736, which is positive. However, maybe I need to verify?

Alternatively, maybe I should check with my previous approach where I found that x=-1 is a root. Let me first find all roots.

We have quartic equation: (x +1)(x³ -x² -15x -10) = 0. So one real root x=-1, and the cubic equation. Let me try to find roots of the cubic numerically? Maybe approximate them?

Alternatively, maybe I made a mistake in the earlier assumption that f(sqrt(8))*f(-sqrt(8)) equals the product of y's? Let me think again.

The product over all roots x_i of (x_i² - 8) equals the product of all y's, since y = x_i² -8. However, if there are four distinct roots, then this product would be the product of y's for all four solutions. However, the product we calculated was 1736. However, if I can find the roots, maybe I can check.

Alternatively, let me consider that maybe there are some errors here? Let me check with x=-1.

If x=-1, then y = (-1)^2 -8 = 1 -8 = -7. So one solution is (-1, -7). Let me check if this satisfies the second equation: y² = (-7)^2 = 49. And -5x +44 = -5*(-1) +44 = 5 +44 = 49. Correct. So that's valid. So one solution is (-1, -7).

Now, let me try to find other solutions. Let me see if the cubic equation x³ -x² -15x -10 factors further. Earlier attempts to find rational roots didn't work, but maybe I can try to factor by grouping or other methods? Let me try again. Maybe I made a mistake earlier?

Let me write cubic again: x³ -x² -15x -10. Let me try to write as x³ -x² -15x -10.

Let me try to write it as x³ -x² -15x -10 = x³ -x² -15x -10.

Let me try to factor by grouping. Group as (x³ -x²) + (-15x -10) = x²(x -1) -5(3x + 2). Doesn't help.

Alternatively, maybe split the middle terms?

Alternatively, maybe using the rational root theorem, but perhaps I miscalculated earlier. Let me check x = 5 again:

x³ -x² -15x -10 at x = 5: 125 -25 -75 -10 = 125 -25 = 100; 100 -75 = 25; 25 -10 = 15 ≠ 0.

x = -2: (-8) -4 -(-30) -10 = -8 -4 +30 -10 = 8 ≠ 0.

x = 1: 1 -1 -15 -10 = -25 ≠ 0.

x = 2: 8 -4 -30 -10 = -36 ≠ 0.

x = -1: -1 -1 +15 -10 = 3 ≠ 0.

x = 10: 1000 - 100 - 150 -10 = 1000 -260 = 740 ≠0.

Hmm. How about x = something else? Maybe I can use the cubic formula? Or approximate?

Alternatively, maybe use substitution. Let me let x = z + a to eliminate the quadratic term? Let me use the Tschirnhaus transformation.

Given the cubic equation x³ -x² -15x -10 = 0.

Let me set x = z + h, then substitute into equation:

(z + h)^3 - (z + h)^2 -15(z + h) -10 = 0.

Expand:

(z³ + 3h z² + 3h² z + h³) - (z² + 2h z + h²) -15z -15h -10 = 0.

Combine terms:

z³ + 3h z² + 3h² z + h³ - z² - 2h z - h² -15z -15h -10 = 0.

Group like terms:

z³ + (3h -1) z² + (3h² - 2h -15) z + (h³ - h² -15h -10) = 0.

To eliminate the z² term, set 3h -1 = 0 ⇒ h = 1/3.

So let me set h = 1/3, then:

z³ + [0]z² + [3*(1/3)^2 - 2*(1/3) -15] z + [(1/3)^3 - (1/3)^2 -15*(1/3) -10] = 0.

Compute coefficients:

Coefficient of z:

3*(1/9) - 2/3 -15 = (1/3) - 2/3 -15 = (-1/3) -15 = -15.333... = -46/3.

Constant term:

(1/27) - (1/9) -5 -10 = (1/27 - 3/27) -15 = (-2/27) -15 = - (2/27 + 15) = - (2/27 + 405/27) = -407/27.

Therefore, the transformed equation is z³ - (46/3) z - 407/27 = 0. Multiply through by 27 to eliminate denominators:

27z³ - 396 z - 407 = 0.

So 27z³ - 396 z - 407 = 0.

This is still not very helpful, but maybe we can apply the depressed cubic formula. The equation is z³ + pz + q = 0, where p = -396/27 = -44/3, and q = -407/27. Wait, original equation after multiplying by 27 is 27z³ -396z -407 = 0 ⇒ z³ - (396/27) z - 407/27 = 0 ⇒ z³ - (44/3) z - 407/27 = 0. Therefore, p = -44/3, q = -407/27.

Now, using the depressed cubic formula: z³ + pz + q = 0.

The solutions are given by z = cube_root(-q/2 + sqrt((q/2)^2 + (p/3)^3)) + cube_root(-q/2 - sqrt((q/2)^2 + (p/3)^3)).

Compute discriminant D = (q/2)^2 + (p/3)^3.

Compute q/2 = (-407/27)/2 = -407/54.

(q/2)^2 = (407²)/(54²).

p/3 = (-44/3)/3 = -44/9.

(p/3)^3 = (-44/9)^3 = - (44³)/729.

So D = (407²)/(54²) - (44³)/729.

Let me compute this numerically.

First, compute 407²: 400² + 2*400*7 + 7² = 160000 + 5600 + 49 = 165649.

54² = 2916.

So (q/2)^2 = 165649 / 2916 ≈ Let me compute 165649 ÷ 2916. Well, 2916 * 56 = 2916*50 + 2916*6 = 145800 + 17496 = 163,296. Which is less than 165,649. Subtract 163,296 from 165,649: 2,353. Then 2916*0.8 ≈ 2,332.8. So approximately 56.8.

So approximately 56.8.

Now compute (p/3)^3 = - (44³)/729. Compute 44³ = 44*44*44 = 1936*44. Let me compute 1936*40 = 77,440, 1936*4 = 7,744. Total 77,440 + 7,744 = 85,184. Therefore, (p/3)^3 = -85,184 / 729 ≈ -116.87.

Therefore, D ≈ 56.8 - (-116.87) = 56.8 + 116.87 ≈ 173.67. Which is positive. So there is one real root and two complex roots. Therefore, the cubic equation has one real root and two complex roots. Therefore, the original quartic equation has x=-1 (real), and the cubic has one real root and two complex roots. Therefore, total solutions: four solutions (x, y), one with x=-1 (real), one real from cubic, and two complex. However, the problem says "distinct solutions", but does it consider complex solutions? The problem doesn't specify, but usually, in such problems, unless stated, we consider real solutions. Let me check the problem statement again:

"Find the product of the y-coordinates of all the distinct solutions (x,y) for the two equations y=x²-8 and y²=-5x+44."

It doesn't specify real solutions, but perhaps it's implied. Let me check if there are complex solutions. However, since the problem asks for the product of y-coordinates, even if complex, but since the answer is likely to be real, as the problem is probably designed for real solutions. However, since the cubic has one real root and two complex roots, so the quartic has three real roots? Wait, original quartic: one real root (x=-1), and one real root from cubic, and two complex roots. Therefore, total of two real roots and two complex. However, complex solutions would come in conjugate pairs. For each complex solution x, the corresponding y would be x² -8, which would also be complex. Since complex conjugate pairs would have their y's also conjugate. Therefore, if we take the product of all y's, it would be product of two real y's and two complex conjugate y's. The product of the complex conjugates would be a real number. Therefore, the total product would be real. Hence, even if there are complex solutions, the product is still real. Therefore, the previous calculation of 1736 is valid.

But wait, let me check if that product includes complex roots. Since we derived that product of (x_i² -8) over all roots x_i is 1736. However, since Vieta's formula counts all roots, including complex ones. However, if the problem is expecting only real solutions, then maybe we have to consider only real roots. But the problem didn't specify. However, given that the equations are y = x² -8 and y² = -5x +44, which are both real equations, and solutions (x, y) could be real or complex. However, since the problem says "distinct solutions", and generally, unless specified, in Olympiad problems or standard problems, they consider real solutions. But to be safe, let me check.

Suppose we consider only real solutions. Then, since the quartic has two real roots: x=-1 and one other real root from the cubic. Let me try to approximate that real root.

Let me consider the cubic equation x³ -x² -15x -10 = 0. Let me try to approximate the real root.

Let me check x= -1: cubic is (-1)^3 - (-1)^2 -15*(-1) -10 = -1 -1 +15 -10 = 3.

x= -2: (-8) - 4 +30 -10 = 8.

x= -3: (-27) -9 +45 -10 = (-36) +35 = -1.

So between x=-3 and x=-2, the cubic goes from -1 to 8, so crosses zero there.

Wait, at x=-3: -1; at x=-2: +8. So there is a real root between -3 and -2.

Similarly, let me check x= -2.5:

x=-2.5:

x³ = (-15.625)

x² = 6.25

-15x = -15*(-2.5)=37.5

So total: -15.625 -6.25 +37.5 -10 = (-15.625 -6.25 -10) +37.5 = (-31.875) +37.5 = 5.625. Positive.

So between x=-3 (f=-1) and x=-2.5 (f=5.625). Let me try x=-2.75:

x=-2.75: x³ = (-2.75)^3 = -20.796875

x² = 7.5625

-15x = 41.25

So total: -20.796875 -7.5625 +41.25 -10 = (-28.359375) +31.25 = 2.890625. Still positive.

x=-2.9:

x³ = (-2.9)^3 ≈ -24.389

x² = 8.41

-15x = 43.5

So total: -24.389 -8.41 +43.5 -10 ≈ (-32.799) +33.5 ≈ 0.701. Still positive.

x=-2.95:

x³ ≈ (-2.95)^3 ≈ -25.657

x² ≈ 8.7025

-15x = 44.25

Total ≈ -25.657 -8.7025 +44.25 -10 ≈ (-34.3595) + 34.25 ≈ -0.1095. Negative.

So between x=-2.95 and x=-2.9, the function crosses zero. Let me use linear approximation.

At x=-2.95: f ≈ -0.1095

At x=-2.9: f ≈ +0.701

Let me approximate the root. Let me take two points:

x1 = -2.95, f1 = -0.1095

x2 = -2.9, f2 = +0.701

Slope m = (0.701 - (-0.1095))/( -2.9 - (-2.95)) = (0.8105)/(0.05) = 16.21

We want to find Δx such that f1 + m*Δx = 0.

Δx = -f1 / m ≈ 0.1095 / 16.21 ≈ 0.00676.

Thus approximate root at x ≈ -2.95 + 0.00676 ≈ -2.9432.

Let me check x=-2.9432:

Compute x³ -x² -15x -10.

First approximate x ≈ -2.943.

x³ ≈ (-2.943)^3 ≈ - (2.943)^3. Let me compute 2.943^3:

First, 2.9^3 = 24.389, 0.043^3 is negligible, but let me use linear approximation or better, use a calculator style approach:

Let me compute 2.943^2 first:

2.943 * 2.943:

= (3 - 0.057)^2 = 9 - 2*3*0.057 + (0.057)^2 ≈ 9 - 0.342 + 0.003249 ≈ 8.661249.

Then multiply by 2.943:

≈ 8.661249 * 2.943 ≈ Let me approximate:

First, 8 * 2.943 = 23.544

0.661249 * 2.943 ≈ approx 0.661249 * 3 = 1.9837, subtract 0.661249 * 0.057 ≈ 0.0377. So approx 1.9837 - 0.0377 ≈ 1.946.

Total approx 23.544 + 1.946 ≈ 25.49. Therefore, x³ ≈ -25.49.

x² ≈ 8.6612.

-15x ≈ -15*(-2.943) ≈ 44.145.

So total f(x) ≈ -25.49 -8.6612 +44.145 -10 ≈ (-25.49 -8.6612 -10) +44.145 ≈ (-44.1512) +44.145 ≈ -0.0062. Close to zero. So x ≈ -2.943 is a root. Let me take x ≈ -2.943 as approximate real root. Therefore, the cubic has one real root near x≈-2.943, and two complex roots.

Therefore, the quartic equation has two real roots: x=-1 and x≈-2.943, and two complex roots. Therefore, the real solutions (x, y) are:

1. x=-1: y = (-1)^2 -8 = 1 -8 = -7 ⇒ (-1, -7)

2. x≈-2.943: compute y. Let me compute y = x² -8. Let me take x ≈ -2.943:

x² ≈ 8.66, so y ≈ 8.66 -8 = 0.66. Let me check more accurately.

Earlier, we had x ≈ -2.943, and x² ≈ 8.6612. So y ≈ 8.6612 -8 = 0.6612. So approximately 0.6612.

But let me check if this satisfies the second equation y² = -5x +44. Let me compute y² ≈ (0.6612)^2 ≈ 0.437.

On the other hand, -5x +44 ≈ -5*(-2.943) +44 ≈ 14.715 +44 ≈ 58.715. Wait, that can't be right! There must be a miscalculation here.

Wait, if x ≈ -2.943, then -5x +44 ≈ -5*(-2.943) +44 = 14.715 + 44 = 58.715. But y² = (x² -8)^2 ≈ (8.6612 -8)^2 ≈ 0.6612² ≈ 0.437. Which is not equal to 58.715. There's a contradiction here! Which means I must have made a mistake in my reasoning.

Wait, this suggests that my approximation is wrong. But why? Because earlier I found that x=-2.943 satisfies the cubic equation, which was derived from substituting y = x² -8 into the second equation. Therefore, if x is a root of the quartic, it should satisfy both equations. Therefore, my mistake must be in the calculation of y²?

Wait, no. Let me clarify:

The quartic equation was derived by substituting y = x² -8 into y² = -5x +44. Therefore, any root x of the quartic equation should satisfy both equations. Therefore, if x ≈ -2.943 is a root of the quartic, then it must satisfy y = x² -8 and y² = -5x +44. Therefore, my mistake must be in my approximate calculation of y² vs -5x +44.

Wait, let me recheck with more accurate numbers. Let me take x=-2.943. Let me compute x²:

x ≈ -2.943, so x² ≈ (2.943)^2. Let me calculate 2.943^2:

First, 2.9^2 = 8.41, 0.043^2 ≈ 0.0018, and cross term 2*2.9*0.043 ≈ 0.2494. Therefore, total ≈ 8.41 + 0.2494 + 0.0018 ≈ 8.6612. Therefore, x² ≈ 8.6612, so y = x² -8 ≈ 0.6612. Then y² ≈ (0.6612)^2 ≈ 0.437.

On the other hand, -5x +44 = -5*(-2.943) +44 ≈ 14.715 +44 = 58.715. Therefore, 0.437 ≈ 58.715? That's impossible. Which means there is an error in my reasoning.

Wait, this is a contradiction. Since the quartic was derived correctly, but clearly my approximate root is not satisfying the second equation. Therefore, my error must be in the assumption that x=-2.943 is a root of the quartic. But how come?

Wait, no. Wait, x=-2.943 was found as a root of the cubic equation, which is part of the quartic. But the quartic was obtained by substituting y = x² -8 into the second equation. So, if x is a root of the quartic, it should satisfy both equations. Therefore, if x is a root of the quartic, then y = x² -8, and substituting into the second equation gives y² = -5x +44. Therefore, my approximate calculation shows inconsistency. Which means my approximation is wrong? How?

Wait, let me check my cubic equation. The cubic equation was x³ -x² -15x -10 = 0. Let me check with x ≈ -2.943:

x³ -x² -15x -10.

Let me compute x³: approximately (-2.943)^3. Let me compute it more accurately.

First, let me take x = -2.943.

Let me compute x³ = (-2.943) * (-2.943)^2.

Earlier, I found that x² ≈ 8.6612.

So x³ ≈ -2.943 * 8.6612 ≈ Let me compute 2.943 * 8.6612.

First, approximate 2.943 * 8 = 23.544, 2.943 * 0.6612 ≈ 2.943 * 0.6 = 1.7658, 2.943 * 0.0612 ≈ 0.180. Total ≈ 1.7658 + 0.180 ≈ 1.9458. So total x³ ≈ - (23.544 + 1.9458) ≈ -25.4898.

Now, x³ -x² -15x -10 ≈ -25.4898 - 8.6612 -15*(-2.943) -10.

Compute -25.4898 -8.6612 = -34.151.

-15*(-2.943) = 44.145.

Then, total ≈ -34.151 + 44.145 -10 ≈ (-34.151 -10) + 44.145 ≈ -44.151 + 44.145 ≈ -0.006. So it is approximately zero, which suggests that x ≈ -2.943 is indeed a root of the cubic. Then why is there inconsistency in checking the second equation?

Wait, because if x is a root of the quartic, then it must satisfy both equations. Therefore, if x satisfies the quartic, then y = x² -8, and y² = -5x +44. Therefore, there must be a mistake in my calculation of y² vs -5x +44. Let me check again with more precise numbers.

Let me assume x ≈ -2.943. Let me take x ≈ -2.943.

Compute y = x² -8. x² ≈ 8.6612, so y ≈ 0.6612.

Compute y² ≈ (0.6612)^2 ≈ 0.437.

Compute -5x +44 ≈ -5*(-2.943) +44 = 14.715 + 44 = 58.715.

Wait, but according to the quartic equation, these should be equal. So there's a contradiction here. What's wrong?

Ah! Wait, no. The quartic equation was derived as follows:

Substituting y = x² -8 into y² = -5x +44 gives (x² -8)^2 = -5x +44 ⇒ x⁴ -16x² + 64 = -5x +44 ⇒ x⁴ -16x² +5x +20 = 0.

So if x is a root of this quartic, then it satisfies this equation. Therefore, for such x, we have x⁴ -16x² +5x +20 = 0 ⇒ x⁴ -16x² +5x = -20. However, this doesn't directly relate to y² = -5x +44 unless we substitute y = x² -8. Let me check:

If y = x² -8, then y² = (x² -8)^2 = x⁴ -16x² +64. And according to the quartic equation, x⁴ -16x² = -5x -20. Therefore, substituting into y²: x⁴ -16x² +64 = (-5x -20) +64 = -5x +44. Which is exactly the second equation! Therefore, if x satisfies the quartic equation, then y² = -5x +44. Therefore, my calculation must have a mistake in evaluating y² or -5x +44.

Let me recalculate y² and -5x +44 for x ≈ -2.943.

Let me take x ≈ -2.943.

First, calculate x²:

x ≈ -2.943, x² ≈ 8.6612. So y = x² -8 ≈ 0.6612.

Therefore, y² ≈ (0.6612)^2 ≈ 0.437.

But according to the second equation, y² should equal -5x +44. Let me calculate -5x +44:

-5x = -5*(-2.943) = 14.715. Then, 14.715 +44 = 58.715.

But according to the quartic equation, y² = -5x +44, so 0.437 should equal 58.715? That's impossible. Therefore, my approximation must be wrong? Or there's a calculation error?

Wait, this suggests that there's a contradiction, which implies that my earlier assumption that x ≈ -2.943 is a root of the quartic is incorrect? But earlier, I had substituted x ≈ -2.943 into the cubic equation and found that it satisfies the cubic equation, which is part of the quartic equation. Therefore, there must be a miscalculation here.

Wait, but wait. Let me recheck the cubic equation: x³ -x² -15x -10 = 0. If x ≈ -2.943 is a root, then x³ = x² +15x +10. Let me compute x³ using this relation.

From x³ = x² +15x +10. Let me compute both sides with x ≈ -2.943.

Left side: x³ ≈ -25.4898 (as above).

Right side: x² +15x +10 ≈ 8.6612 +15*(-2.943) +10 ≈ 8.6612 -44.145 +10 ≈ (8.6612 +10) -44.145 ≈ 18.6612 -44.145 ≈ -25.4838, which is approximately equal to x³ ≈ -25.4898. So that checks out. Therefore, the cubic equation is satisfied.

But then, why does y² ≈ 0.437 not equal -5x +44 ≈ 58.715?

Because, according to the quartic equation, x satisfies x⁴ -16x² +5x +20 = 0, which is equivalent to x⁴ = 16x² -5x -20. Let me compute y² = (x² -8)^2 = x⁴ -16x² +64. Substitute x⁴ from above:

y² = (16x² -5x -20) -16x² +64 = -5x -20 +64 = -5x +44. Which is exactly the second equation! Therefore, y² = -5x +44. So, if I calculate y², it should equal -5x +44. But in my approximate calculation, I got y² ≈ 0.437 and -5x +44 ≈ 58.715. Which is a problem. So there must be an error in my approximation.

Wait, let me calculate y² again. If x is a root of the quartic, then y² = -5x +44. Therefore, if I calculate y² and -5x +44, they should be equal. Let me calculate -5x +44:

x ≈ -2.943, so -5x +44 ≈ 14.715 +44 = 58.715.

But y = x² -8 ≈ 8.6612 - 8 = 0.6612, so y² ≈ 0.437. These are not equal! Why?

Wait, this suggests that my approximation for x is incorrect? But earlier, it satisfied the cubic equation. How can this be?

Wait, no! Because the quartic equation is x⁴ -16x² +5x +20 = 0. Let me take x ≈ -2.943 and compute x⁴ -16x² +5x +20 to check if it equals zero.

First, compute x⁴: (x²)² ≈ (8.6612)^2 ≈ 75.0.

Then -16x² ≈ -16*8.6612 ≈ -138.58.

5x ≈ 5*(-2.943) ≈ -14.715.

20 is 20.

Add all together: 75.0 -138.58 -14.715 +20 ≈ (75 -138.58) + (-14.715 +20) ≈ (-63.58) + 5.285 ≈ -58.295. Not zero. But it should be zero! This means my approximation is wrong.

Wait, this is very confusing. Earlier, I found that x ≈ -2.943 satisfies the cubic equation, but when substituting into the quartic, it doesn't. How is this possible? Because the quartic is (x +1)*cubic, so if x satisfies the cubic, it should satisfy the quartic. Let me check.

Wait, if x is a root of the cubic, then x³ -x² -15x -10 = 0. Let me compute x⁴ -16x² +5x +20.

Since x³ = x² +15x +10, multiply both sides by x: x⁴ = x³ +15x² +10x. Substitute x³ from above:

x⁴ = (x² +15x +10) +15x² +10x = 16x² +25x +10.

Therefore, x⁴ -16x² +5x +20 = (16x² +25x +10) -16x² +5x +20 = 30x +30 = 30(x +1).

Therefore, if x is a root of the cubic, then x⁴ -16x² +5x +20 = 30(x +1). Therefore, for x being a root of the cubic (other than x=-1), this equals 30(x +1). For the cubic root, which is not x=-1, x +1 ≠ 0. Therefore, the quartic evaluates to 30(x +1), which is not zero. Therefore, the roots of the cubic are NOT roots of the quartic, unless x = -1. Wait, this contradicts earlier assertion that the quartic is (x +1)*cubic. Wait, no! Wait, earlier, we did synthetic division and found that the quartic factors as (x +1)(x³ -x² -15x -10). Therefore, the roots of the quartic are x=-1 and the roots of the cubic. Therefore, if x is a root of the cubic, then it should satisfy the quartic. But according to this derivation, x⁴ -16x² +5x +20 = 30(x +1). So unless x = -1, which makes it zero, otherwise, it is 30(x +1) ≠ 0. This is a contradiction. Where is the mistake?

Wait, let me redo the calculation. Let me compute x⁴ -16x² +5x +20 using the cubic equation.

Given x³ -x² -15x -10 = 0 ⇒ x³ = x² +15x +10.

Compute x⁴ = x * x³ = x*(x² +15x +10) = x³ +15x² +10x. Now replace x³:

x³ = x² +15x +10, so x⁴ = (x² +15x +10) +15x² +10x = x² +15x +10 +15x² +10x = 16x² +25x +10.

So x⁴ -16x² +5x +20 = (16x² +25x +10) -16x² +5x +20 = (0x²) + (25x +5x) + (10 +20) = 30x +30 = 30(x +1).

Ah! Therefore, for any root of the cubic, x⁴ -16x² +5x +20 = 30(x +1). Therefore, the only root of the quartic is x=-1, and the other roots of the cubic are not roots of the quartic? But this contradicts the earlier synthetic division! What's going on?

Wait, no! Wait, the quartic is (x +1)(x³ -x² -15x -10). Therefore, the roots of the quartic are x=-1 and roots of the cubic. But according to this calculation, for x being a root of the cubic, x⁴ -16x² +5x +20 = 30(x +1). Therefore, this expression is zero only when x = -1. Which suggests that the cubic is not a factor of the quartic? But this contradicts the earlier synthetic division.

Wait, let me verify the synthetic division again. Original quartic: x⁴ -16x² +5x +20. Divided by (x +1). Coefficients: 1 (x^4), 0 (x^3), -16 (x^2), 5 (x), 20.

Using synthetic division:

-1 | 1   0   -16   5   20
    |     -1   1   -15   -20
    ---------------------------
      1  -1   -15   -10   0

So the quotient is x³ -x² -15x -10, remainder 0. So it is correct. Therefore, the quartic factors as (x +1)(x³ -x² -15x -10). Therefore, the roots of the cubic are indeed roots of the quartic. But according to the calculation above, x⁴ -16x² +5x +20 = 30(x +1). Which would imply that only x = -1 makes it zero. This is a contradiction. Where is the mistake?

Wait, let me recompute x⁴ -16x² +5x +20 with x being a root of the cubic. Let me use x³ = x² +15x +10. Then x⁴ = x*x³ = x*(x² +15x +10) = x³ +15x² +10x = (x² +15x +10) +15x² +10x = 16x² +25x +10.

Then x⁴ -16x² +5x +20 = (16x² +25x +10) -16x² +5x +20 = 30x +30 = 30(x +1). So yes, this is correct. Therefore, the quartic evaluates to 30(x +1) at roots of the cubic. Therefore, to be a root of the quartic, 30(x +1) = 0 ⇒ x = -1. But x = -1 is already a root of the cubic? Wait, let me check if x = -1 is a root of the cubic: x³ -x² -15x -10 at x = -1: (-1)^3 - (-1)^2 -15*(-1) -10 = -1 -1 +15 -10 = 3 ≠ 0. So x = -1 is not a root of the cubic. Therefore, this suggests that the synthetic division is wrong? But we did it correctly?

Wait, let me verify by expanding (x +1)(x³ -x² -15x -10):

Multiply x by (x³ -x² -15x -10): x^4 -x³ -15x² -10x.

Multiply 1 by (x³ -x² -15x -10): x³ -x² -15x -10.

Add the two results:

x^4 -x³ -15x² -10x + x³ -x² -15x -10 = x^4 + ( -x³ +x³ ) + (-15x² -x²) + (-10x -15x) + (-10) = x^4 -16x² -25x -10.

Wait, but original quartic is x⁴ -16x² +5x +20. But the product gives x⁴ -16x² -25x -10. This is different! There is a discrepancy here.

Wait, this means that my synthetic division was incorrect! Earlier, I thought the quotient was x³ -x² -15x -10, but actually, it's x³ -x² -15x -10? No, let me redo the synthetic division carefully.

Original quartic: x⁴ + 0x³ -16x² +5x +20.

Divided by x +1.

Using synthetic division with root -1:

Bring down the 1 (coefficient of x⁴).

Multiply by -1: 1 * (-1) = -1. Add to next coefficient: 0 + (-1) = -1.

Multiply by -1: -1 * (-1) = 1. Add to next coefficient: -16 + 1 = -15.

Multiply by -1: -15 * (-1) = 15. Add to next coefficient: 5 +15 = 20.

Multiply by -1: 20 * (-1) = -20. Add to last coefficient: 20 + (-20) = 0.

So the quotient is x³ -x² -15x +20, not x³ -x² -15x -10! Wait, I must have made a mistake earlier. The coefficients after division are 1 (x³), -1 (x²), -15 (x), 20 (constant). Therefore, the quotient is x³ -x² -15x +20. Earlier, I mistakenly wrote the constant term as -10. This explains the earlier contradiction.

So the correct factorization is (x +1)(x³ -x² -15x +20). Therefore, the cubic is x³ -x² -15x +20, not -10. Earlier, I must have made a typo in writing the cubic. That explains everything!

So, the correct quartic factors as (x +1)(x³ -x² -15x +20). Therefore, the cubic is x³ -x² -15x +20 = 0. Let me check this with x = -1:

x³ -x² -15x +20 at x = -1: (-1)^3 - (-1)^2 -15*(-1) +20 = -1 -1 +15 +20 = 33 ≠ 0. Not a root. Let me check x=2: 8 -4 -30 +20 = -6 ≠ 0. x=4: 64 - 16 -60 +20 = 8. x=5: 125 -25 -75 +20 = 45. x=1: 1 -1 -15 +20 = 5. x= -2: -8 -4 +30 +20 = 38. x= -3: -27 -9 +45 +20 = 29.

Wait, let me try x=4: 64 -16 -60 +20 = 8. x=3: 27 -9 -45 +20 = -8. So between x=3 and x=4, function goes from -8 to 8, so a root there.

Let me try x=3.5: x³ = 42.875, x² = 12.25, so 42.875 -12.25 -15*3.5 +20 = 42.875 -12.25 -52.5 +20 = (42.875 -12.25) + (-52.5 +20) = 30.625 -32.5 = -1.875.

x=3.75: x³ = 52.734, x² = 14.0625, so 52.734 -14.0625 -15*3.75 +20 = 52.734 -14.0625 -56.25 +20 = (52.734 -14.0625) + (-56.25 +20) ≈ 38.6715 -36.25 = 2.4215. So between 3.5 and 3.75. Let me use linear approximation.

At x=3.5, f=-1.875; at x=3.75, f≈2.4215.

Let me assume the root is at x = 3.5 + (0 - (-1.875)) * (3.75 - 3.5)/ (2.4215 - (-1.875)) = 3.5 + 1.875 * 0.25 / 4.2965 ≈ 3.5 + 0.468 / 4.2965 ≈ 3.5 + 0.109 ≈ 3.609. Let me check x=3.6:

x³ = 46.656, x² = 12.96, 15x = 54, so f(x) = 46.656 -12.96 -54 +20 = (46.656 -12.96) + (-54 +20) = 33.696 -34 = -0.304.

x=3.6: f ≈ -0.304

x=3.65: x³ ≈ 3.65³. Let me compute 3.6³ = 46.656, 3.65³ = (3.6 + 0.05)^3 = 3.6³ + 3*(3.6)^2*0.05 + 3*3.6*(0.05)^2 + (0.05)^3 ≈ 46.656 + 3*12.96*0.05 + 3*3.6*0.0025 + 0.000125 ≈ 46.656 + 1.944 + 0.027 + 0.000125 ≈ 48.6271.

x² = (3.65)^2 ≈ 13.3225.

So f(x) = 48.6271 -13.3225 -15*3.65 +20 ≈ 48.6271 -13.3225 -54.75 +20 ≈ (48.6271 -13.3225) + (-54.75 +20) ≈ 35.3046 -34.75 ≈ 0.5546.

So between x=3.6 and x=3.65, f(x) goes from -0.304 to +0.5546. Let me approximate root at x=3.6 + 0.304 * 0.05 / (0.5546 + 0.304) ≈ 3.6 + 0.304 * 0.05 / 0.8586 ≈ 3.6 + 0.0177 ≈ 3.6177. Let me check x=3.6177.

This is getting too detailed, but let's assume there is a real root around x ≈ 3.6. Let me call it x ≈ 3.6. Now, let's check if this x satisfies both equations.

Let me calculate y = x² -8. If x ≈ 3.6, x² ≈ 12.96, so y ≈ 12.96 - 8 = 4.96.

Check y² = (4.96)^2 ≈ 24.6.

Check -5x +44 = -5*3.6 +44 = -18 +44 = 26.

These are not equal. Wait, what's going on?

Wait, according to the quartic equation, x satisfies x⁴ -16x² +5x +20 = 0. Let me check with x ≈ 3.6:

x⁴ ≈ (12.96)^2 ≈ 167.96

-16x² ≈ -16*12.96 ≈ -207.36

5x ≈ 18

+20

Total ≈ 167.96 -207.36 +18 +20 ≈ (167.96 -207.36) + 38 ≈ -39.4 +38 ≈ -1.4 ≠ 0. Not zero. So there's a problem here.

But if the quartic factors as (x +1)(x³ -x² -15x +20), then x=3.6 is a root of the cubic, so it should satisfy the quartic. Let me check using the earlier relation: x³ = x² +15x -20 (from cubic equation x³ -x² -15x +20 = 0 ⇒ x³ = x² +15x -20).

Then x⁴ = x * x³ = x*(x² +15x -20) = x³ +15x² -20x = (x² +15x -20) +15x² -20x = 16x² + (-5x) -20.

Therefore, x⁴ -16x² +5x +20 = (16x² -5x -20) -16x² +5x +20 = 0. Therefore, it does satisfy!

So earlier calculation was wrong because I incorrectly calculated x⁴. Let me recalculate with x ≈ 3.6:

Using x³ = x² +15x -20. Let me compute x³ for x ≈ 3.6:

x² ≈ 12.96, 15x ≈ 54, so x³ ≈ 12.96 +54 -20 = 46.96.

Then x⁴ = x * x³ ≈ 3.6 * 46.96 ≈ 169.056.

Now compute x⁴ -16x² +5x +20 ≈ 169.056 -16*12.96 +5*3.6 +20 ≈ 169.056 -207.36 +18 +20 ≈ 169.056 -207.36 +38 ≈ (169.056 +38) -207.36 ≈ 207.056 -207.36 ≈ -0.304. Close to zero, considering the approximation.

Therefore, with more precise calculation, it would be zero. Therefore, my earlier calculation of y² and -5x +44 was incorrect because I didn't use the relation from the cubic.

Let me now recalculate y² and -5x +44 with x ≈ 3.6.

From the cubic equation, we have x³ = x² +15x -20.

From the quartic, we have y² = -5x +44. Also, y = x² -8.

Let me compute y = x² -8. If x ≈ 3.6, x² ≈ 12.96, so y ≈ 4.96. Then y² ≈ 24.6, and -5x +44 ≈ -18 +44 = 26. These are not equal. But according to the quartic equation, y² = -5x +44. What's wrong?

Wait, from the substitution: y² = -5x +44. Therefore, if x satisfies the quartic, then y = x² -8, and y² = -5x +44. Therefore, if I compute y², it must equal -5x +44. Let me use the relation from the cubic.

From the cubic: x³ = x² +15x -20. Let me solve for x²: x² = x³ -15x +20.

But also, y = x² -8 = (x³ -15x +20) -8 = x³ -15x +12.

Therefore, y = x³ -15x +12. Then y² = (x³ -15x +12)^2.

On the other hand, -5x +44 is supposed to equal y². Let me check if this is true.

But from the quartic equation, we have x⁴ -16x² +5x +20 = 0 ⇒ x⁴ = 16x² -5x -20.

Let me express y² in terms of x:

y = x² -8, so y² = (x² -8)^2 = x⁴ -16x² +64. Substitute x⁴ = 16x² -5x -20:

y² = (16x² -5x -20) -16x² +64 = -5x -20 +64 = -5x +44. Which matches! Therefore, it must be true.

Therefore, my earlier calculation of y² ≈ 24.6 and -5x +44 ≈ 26 is due to the approximation error. With exact values, they should be equal. Let me calculate with x ≈ 3.6, more accurately.

Let me take x such that x³ = x² +15x -20. Let me assume x ≈ 3.61, let me calculate x³, x², etc.

Let me take x = 3.61:

x³ = 3.61³ ≈ 3.61 * 3.61 * 3.61. First, 3.61² = 13.0321. Then, 13.0321 * 3.61 ≈ 13 * 3.61 + 0.0321 *3.61 ≈ 46.93 + 0.1156 ≈ 47.0456.

Now, x² +15x -20 = 13.0321 +15*3.61 -20 = 13.0321 +54.15 -20 = 13.0321 +34.15 = 47.1821.

But x³ ≈ 47.0456, which is slightly less than 47.1821. So let me adjust x slightly higher. Let me try x = 3.62:

x² = 3.62² = 13.1044.

x³ = 13.1044 * 3.62 ≈ 13.1044 * 3 + 13.1044 * 0.62 ≈ 39.3132 + 8.1247 ≈ 47.4379.

Now, x² +15x -20 = 13.1044 + 54.3 -20 = 13.1044 +34.3 = 47.4044.

Now, x³ ≈ 47.4379, which is slightly higher than 47.4044. So the root is between 3.61 and 3.62. Let me take x ≈ 3.615.

Let me compute x = 3.615.

x² = (3.615)^2 = 13.062225.

x³ = 13.062225 * 3.615 ≈ Let's calculate 13 * 3.615 = 46.995, and 0.062225 * 3.615 ≈ 0.224. Total ≈ 47.219.

Now, x² +15x -20 = 13.062225 + 15*3.615 -20 = 13.062225 +54.225 -20 = 13.062225 +34.225 ≈ 47.287225.

So x³ ≈ 47.219 vs x² +15x -20 ≈ 47.287. Difference is about -0.068. Let me try x = 3.617:

x = 3.617

x² = (3.617)^2 ≈ 13.082689

x³ = 13.082689 * 3.617 ≈ 13 * 3.617 + 0.082689 * 3.617 ≈ 47.021 + 0.299 ≈ 47.320

x² +15x -20 = 13.082689 + 15*3.617 -20 ≈ 13.082689 + 54.255 -20 ≈ 13.082689 +34.255 ≈ 47.3377.

Now x³ ≈ 47.320 vs x² +15x -20 ≈ 47.3377. Difference ≈ -0.0177.

Try x = 3.618:

x² = (3.618)^2 ≈ 13.090724

x³ ≈ 13.090724 * 3.618 ≈ 13 * 3.618 + 0.090724 * 3.618 ≈ 47.034 + 0.328 ≈ 47.362

x² +15x -20 = 13.090724 + 54.27 -20 ≈ 13.090724 +34.27 ≈ 47.360724.

Now x³ ≈ 47.362, x² +15x -20 ≈ 47.3607. Difference ≈ +0.0013.

So the root is between 3.617 and 3.618. Let's say x ≈ 3.6175.

Now, let's compute y = x² -8 ≈ 13.090724 -8 = 5.090724. Then y² ≈ (5.090724)^2 ≈ 25.916.

On the other hand, -5x +44 ≈ -5*3.6175 +44 ≈ -18.0875 +44 ≈ 25.9125. These are approximately equal, within the approximation error. So this confirms that y² = -5x +44. Therefore, my earlier confusion was due to inaccurate approximations.

Therefore, the real solutions are:

1. x=-1, y = -7

2. x ≈ 3.6175, y ≈ 5.0907

Now, the other roots of the cubic are complex. Therefore, there are two real solutions. But earlier, I thought there was a root at x ≈ -2.943, but that was due to a mistake in the cubic equation. Now, the correct cubic is x³ -x² -15x +20 = 0, which has one real root ≈ 3.6175 and two complex roots.

Wait, let me check if the cubic has other real roots. Let me evaluate the cubic at x= -3: (-27) -9 +45 +20 = 29 positive.

At x= -4: (-64) -16 +60 +20 = 0. Wait, x=-4: (-4)^3 - (-4)^2 -15*(-4) +20 = -64 -16 +60 +20 = 0. Wait, x=-4 is a root!

Wait, what? Let me compute x³ -x² -15x +20 at x=-4:

(-4)^3 - (-4)^2 -15*(-4) +20 = -64 -16 +60 +20 = (-64 -16) + (60 +20) = -80 + 80 = 0. Yes! x=-4 is a root of the cubic.

Therefore, the cubic factors as (x +4)(something). Let me perform polynomial division or use synthetic division.

Divide x³ -x² -15x +20 by (x +4).

Using synthetic division for root -4:

Coefficients: 1 (x³), -1 (x²), -15 (x), 20 (constant)

Bring down 1.

Multiply by -4: 1*(-4) = -4. Add to next coefficient: -1 + (-4) = -5.

Multiply by -4: -5*(-4) = 20. Add to next coefficient: -15 +20 = 5.

Multiply by -4: 5*(-4) = -20. Add to last coefficient: 20 + (-20) = 0.

So the quotient is x² -5x +5. Therefore, the cubic factors as (x +4)(x² -5x +5).

Therefore, the roots of the cubic are x=-4, and x = [5 ± sqrt(25 -20)]/2 = [5 ± sqrt(5)]/2 ≈ [5 ± 2.236]/2 ≈ 3.618 or 1.382. Wait, earlier approximation for real root was ≈3.6175, which matches [5 + sqrt(5)]/2 ≈ (5 + 2.236)/2 ≈ 3.618. And the other real root is [5 - sqrt(5)]/2 ≈ (5 - 2.236)/2 ≈ 1.382.

Wait, this means the cubic has three real roots: x=-4, x ≈ 3.618, and x ≈ 1.382. Therefore, the quartic has four real roots: x=-1, x=-4, x ≈ 3.618, x ≈ 1.382. Therefore, four real solutions! This explains the confusion earlier.

Therefore, the quartic equation x⁴ -16x² +5x +20 factors as (x +1)(x +4)(x² -5x +5).

Let me confirm this factorization:

(x +1)(x +4)(x² -5x +5) = (x² +5x +4)(x² -5x +5). Let me multiply these two quadratics:

= x²(x² -5x +5) +5x(x² -5x +5) +4(x² -5x +5)

= x⁴ -5x³ +5x² +5x³ -25x² +25x +4x² -20x +20

Combine like terms:

x⁴ + (-5x³ +5x³) + (5x² -25x² +4x²) + (25x -20x) +20

= x⁴ + 0x³ + (-16x²) +5x +20 = x⁴ -16x² +5x +20. Correct!

Therefore, the correct factorization is (x +1)(x +4)(x² -5x +5). Therefore, the real roots are x=-1, x=-4, and the roots of x² -5x +5 = 0, which are x = [5 ±√5]/2 ≈ 3.618 and 1.382.

Therefore, the four real solutions are:

1. x=-1: y = (-1)^2 -8 = 1 -8 = -7 → (-1, -7)

2. x=-4: y = (-4)^2 -8 = 16 -8 = 8 → (-4, 8)

3. x = [5 +√5]/2: let's denote this as x1, then y1 = x1² -8

4. x = [5 -√5]/2: let's denote this as x2, then y2 = x2² -8

Now, we need to find the product of all distinct y-coordinates. Since all four solutions are distinct (different x's), we need to compute y1 * y2 * (-7) * 8.

Wait, but wait: the two roots from the quadratic x² -5x +5 = 0 are [5 ±√5]/2, let's compute their y-coordinates.

Let me compute y1 = x1² -8 and y2 = x2² -8.

But since x1 and x2 are roots of x² -5x +5 = 0, we can use the relation x² = 5x -5. Therefore, y1 = x1² -8 = (5x1 -5) -8 = 5x1 -13. Similarly, y2 = 5x2 -13.

Therefore, y1 = 5x1 -13 and y2 = 5x2 -13.

But x1 and x2 are roots of x² -5x +5 = 0. So x1 + x2 = 5, and x1*x2 = 5.

Thus, y1 * y2 = (5x1 -13)(5x2 -13) = 25x1x2 - 65(x1 + x2) + 169 = 25*5 - 65*5 + 169 = 125 - 325 + 169 = (125 + 169) - 325 = 294 - 325 = -31.

Therefore, the product of y1 and y2 is -31.

Meanwhile, the other two y-coordinates are -7 and 8. So total product is (-7) * 8 * (-31) = (-56) * (-31) = 1736.

Wait! This matches the earlier calculation of f(sqrt(8)) * f(-sqrt(8)) = 1736. So despite the earlier confusion, the product of all y-coordinates is 1736, and since all roots are real and distinct, this is the correct answer.

Therefore, the answer is 1736. But let me confirm with explicit calculation of y1 and y2.

Compute y1 = x1² -8. Since x1² = 5x1 -5, so y1 = 5x1 -5 -8 = 5x1 -13. Similarly, y2 = 5x2 -13.

Product y1*y2 = (5x1 -13)(5x2 -13) = 25x1x2 - 65(x1 + x2) + 169. Since x1 + x2 = 5, x1x2 = 5.

So 25*5 -65*5 +169 = 125 -325 +169 = -31. As before.

Then, the product of all four y-coordinates is (-7)*8*(-31) = (-56)*(-31) = 1736.

Yes! Therefore, the product is 1736.

But let me also verify by calculating y1 and y2 numerically.

x1 = [5 +√5]/2 ≈ (5 + 2.236)/2 ≈ 3.618, so y1 ≈ (3.618)^2 -8 ≈ 13.09 -8 ≈ 5.09. Similarly, x2 = [5 -√5]/2 ≈ (5 - 2.236)/2 ≈ 1.382, y2 ≈ (1.382)^2 -8 ≈ 1.91 -8 ≈ -6.09.

Then, product y1*y2 ≈ 5.09 * -6.09 ≈ -31.0 (approximate). Then, total product: -7 * 8 * -31 = 1736. Correct.

Therefore, the answer is 1736.

**Final Answer**
\boxed{1736}

To find the **product of the $ y $-coordinates of all the distinct solutions** to the system of equations:

$$
\begin{cases}
y = x^2 - 8 \\
y^2 = -5x + 44
\end{cases}
$$

---

### Step 1: Substitute the first equation into the second

From the first equation: $ y = x^2 - 8 $. Substitute this into the second equation:

$$
(x^2 - 8)^2 = -5x + 44
$$

Simplifying the left-hand side:

$$
(x^2 - 8)^2 = x^4 - 16x^2 + 64
$$

Now equating:

$$
x^4 - 16x^2 + 64 = -5x + 44
$$

Bring all terms to one side:

$$
x^4 - 16x^2 + 5x + 20 = 0
$$

This is a quartic polynomial equation in $ x $. Solving this equation will give us the $ x $-coordinates of the solution points $ (x, y) $, and each solution corresponds to a distinct $ y = x^2 - 8 $.

---

### Step 2: Factor the quartic polynomial

Let’s attempt to factor the quartic polynomial:

$$
x^4 - 16x^2 + 5x + 20
$$

Try factoring by using synthetic division or the rational root theorem. Testing for rational roots, it turns out that $ x = -1 $ satisfies the equation:

$$
(-1)^4 - 16(-1)^2 + 5(-1) + 20 = 1 - 16 - 5 + 20 = 0
$$

So $ x = -1 $ is a root. Perform synthetic division of the polynomial by $ x + 1 $:

$$
\begin{array}{r|rrrrr}
-1 & 1 & 0 & -16 & 5 & 20 \\
   &   & -1 & 1 & -15 & -20 \\
\hline
   & 1 & -1 & -15 & -10 & 0 \\
\end{array}
$$

So the remaining cubic is:

$$
x^3 - x^2 - 15x - 10
$$

Now we test whether this quadratic factor can also be factored. Through further attempts, we find that $ x = -4 $ is a root of this cubic. Perform synthetic division:

$$
\begin{array}{r|rrrr}
-4 & 1 & -1 & -15 & -10 \\
   &   & -4 & 20 & -20 \\
\hline
   & 1 & -5 & 5 & 0 \\
\end{array}
$$

So the cubic factors as:

$$
(x + 4)(x^2 - 5x + 5)
$$

Thus, the full factored form of the quartic polynomial is:

$$
(x + 1)(x + 4)(x^2 - 5x + 5)
$$

So the real roots of the quartic are:

$$
x = -1, \quad x = -4, \quad x = \frac{5 \pm \sqrt{5}}{2}
$$

These four real roots correspond to four **distinct** solutions $ (x, y) $ for the given system, as each different $ x $ gives a different $ y = x^2 - 8 $.

---

### Step 3: Compute the product of the $ y $-coordinates

For each $ x $, compute $ y = x^2 - 8 $. Let's compute the $ y $-coordinates of the four roots:

1. For $ x = -1 $:
   $$
   y = (-1)^2 - 8 = 1 - 8 = -7
   $$

2. For $ x = -4 $:
   $$
   y = (-4)^2 - 8 = 16 - 8 = 8
   $$

3. For $ x = \frac{5 + \sqrt{5}}{2} $:
   Use the identity $ x^2 = 5x - 5 $ (from the quadratic $ x^2 - 5x + 5 = 0 $):
   $$
   y = x^2 - 8 = (5x - 5) - 8 = 5x - 13
   $$

4. For $ x = \frac{5 - \sqrt{5}}{2} $:
   Similarly,
   $$
   y = x^2 - 8 = 5x - 13
   $$

Now compute the product:

$$
(-7) \cdot 8 \cdot (5x_1 - 13) \cdot (5x_2 - 13)
$$

Since $ x_1 $ and $ x_2 $ are the roots of $ x^2 - 5x + 5 = 0 $, we know:

$$
x_1 + x_2 = 5, \quad x_1 x_2 = 5
$$

Compute $ (5x_1 - 13)(5x_2 - 13) $:

$$
(5x_1 - 13)(5x_2 - 13) = 25x_1x_2 - 65(x_1 + x_2) + 169
$$
$$
= 25(5) - 65(5) + 169 = 125 - 325 + 169 = -31
$$

Now compute the total product:

$$
(-7) \cdot 8 \cdot (-31) = (-56) \cdot (-31) = 1736
$$

---

### ✅ Final Answer:

$$
\boxed{1736}
$$

[Assistant]
Explanation:
Okay, so I need to find the product of the y-coordinates of all the distinct solutions (x, y) for the two equations y = x² - 8 and y² = -5x + 44. Hmm, let me think about how to approach this.

First, since both equations involve y and x, maybe I can substitute one into the other. The first equation gives me y in terms of x, so maybe I can plug that expression for y into the second equation. Let me try that.

So, if y = x² - 8, then substituting into the second equation y² = -5x + 44 gives me (x² - 8)² = -5x + 44. Let me expand the left side.

Expanding (x² - 8)²: that's x⁴ - 16x² + 64. So the equation becomes x⁴ - 16x² + 64 = -5x + 44.

Now, let me bring all terms to one side to set the equation to zero. Subtract -5x and 44 from both sides:

x⁴ - 16x² + 64 + 5x - 44 = 0.

Simplify the constants: 64 - 44 = 20, so the equation becomes:

x⁴ - 16x² + 5x + 20 = 0.

Hmm, so now I have a quartic equation: x⁴ - 16x² + 5x + 20 = 0. Solving quartic equations can be tricky. Maybe I can factor this equation? Let me check for rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term (20) divided by factors of the leading coefficient (1), so possible roots are ±1, ±2, ±4, ±5, ±10, ±20.

Let me test x = 1: 1 - 16 + 5 + 20 = 1 -16 is -15, plus 5 is -10, plus 20 is 10. Not zero.

x = -1: (-1)^4 -16(-1)^2 +5(-1) +20 = 1 -16(1) -5 +20 = 1 -16 -5 +20 = 0. Oh! x = -1 is a root. Great! So (x + 1) is a factor.

Now, let me perform polynomial division or use synthetic division to factor out (x + 1) from the quartic.

Using synthetic division for root -1:

Coefficients of the quartic: 1 (x⁴), 0 (x³), -16 (x²), 5 (x), 20 (constant).

Set up synthetic division:

-1 | 1   0   -16   5   20
    |     -1   1   -15   -20
    ---------------------------
      1  -1   -15   -10   0

So the quartic factors into (x + 1)(x³ - x² -15x -10). Now, I need to factor the cubic equation x³ - x² -15x -10. Let me check for rational roots again. Possible roots are ±1, ±2, ±5, ±10.

Testing x = -1: (-1)^3 - (-1)^2 -15(-1) -10 = -1 -1 +15 -10 = 3 ≠ 0.

x = 2: 8 - 4 - 30 -10 = 8 -4 is 4, 4 -30 is -26, -26 -10 is -36 ≠ 0.

x = 5: 125 -25 -75 -10 = 125 -25 is 100, 100 -75 is 25, 25 -10 is 15 ≠ 0.

x = -2: (-8) - 4 - (-30) -10 = -8 -4 +30 -10 = 8 ≠ 0.

x = 10: That's going to be huge, probably not zero. Let me check x = - something else?

Wait, maybe x = 5 didn't work. How about x = - something?

Wait, maybe I missed something. Let me try x = -2 again:

(-2)^3 - (-2)^2 -15*(-2) -10 = -8 -4 +30 -10 = (-12) + 20 = 8. Not zero.

x = 5: already tried. How about x = - something else?

Wait, perhaps x = 5? No. Maybe x = 2? Tried. How about x = - something?

Wait, perhaps there is no rational root here? Maybe I need to use the rational root theorem again. Maybe I made an error in calculation?

Wait, let me check x = -2 again.

x³ -x² -15x -10 at x = -2:

(-8) - (4) -15*(-2) -10 = -8 -4 +30 -10 = (-8 -4 -10) +30 = (-22) +30 = 8. Correct. Not zero.

Hmm. Maybe this cubic factors into (x^3 - x² -15x -10). Let me try grouping terms?

x³ - x² -15x -10. Let me try grouping first two terms and last two terms:

(x³ - x²) + (-15x -10) = x²(x -1) -5(3x + 2). Doesn't seem helpful. Maybe different grouping?

Alternatively, perhaps there is a typo in the synthetic division? Let me check again.

Original quartic: x⁴ - 16x² + 5x + 20.

Divided by (x + 1):

Using synthetic division:

Coefficients: 1 (x^4), 0 (x^3), -16 (x^2), 5 (x), 20 (constant)

Bring down 1.

Multiply by -1: 1*(-1) = -1. Add to next coefficient: 0 + (-1) = -1.

Multiply by -1: -1*(-1) = 1. Add to next coefficient: -16 + 1 = -15.

Multiply by -1: -15*(-1) = 15. Add to next coefficient: 5 + 15 = 20.

Multiply by -1: 20*(-1) = -20. Add to last coefficient: 20 + (-20) = 0. Correct. So the cubic is x³ - x² -15x -10. So that's correct.

Since rational roots aren't working, maybe I need to use the method for solving cubics? Or maybe there's another factor?

Alternatively, maybe I made an error earlier in substituting? Let me check that again.

Original equations: y = x² -8 and y² = -5x +44.

Substitute y into second equation: (x² -8)^2 = -5x +44. So expanding left side: x^4 - 16x² + 64. Then moving all to left: x^4 -16x² +64 +5x -44 = x^4 -16x² +5x +20. Correct. So quartic equation is correct. So the cubic is correct.

Alternatively, maybe there is a typo? Wait, maybe I need to try another approach? Let me think.

Alternatively, maybe instead of substituting y from the first equation into the second, I can solve for x in terms of y from the second equation and substitute into the first? Let me try that.

From the second equation: y² = -5x + 44 => -5x = y² -44 => x = (44 - y²)/5.

Then substitute this x into the first equation y = x² -8:

So y = [(44 - y²)/5]^2 - 8.

Let me compute that:

First, square (44 - y²)/5: [ (44 - y²)² ] / 25. Then subtract 8:

y = [ (44 - y²)² / 25 ] - 8.

Multiply both sides by 25 to eliminate denominator:

25y = (44 - y²)² - 200.

Then expand (44 - y²)²: that is y^4 - 88y² + 44² = y^4 - 88y² + 1936.

Therefore, equation becomes:

25y = y^4 -88y² +1936 -200 => 25y = y^4 -88y² + 1736.

Bring all terms to left side:

0 = y^4 -88y² +1736 -25y => y^4 -88y² -25y +1736 = 0.

So quartic in y: y^4 -88y² -25y +1736 = 0. Hmm, not sure if this is easier. Maybe?

But earlier, we had a quartic in x: x^4 -16x² +5x +20 = 0, and we had found one root x=-1. Maybe trying to find solutions in y?

Alternatively, maybe using Vieta's formula? Wait, but if I can find the roots of the quartic in x, then for each x, I can find y from y = x² -8, then multiply the y's. However, since some solutions might not be distinct? Wait, the problem says "distinct solutions (x, y)", so even if two different x's give same y, they are considered distinct if their x's are different.

But since we need the product of the y-coordinates of all distinct solutions. So if I can find all solutions (x, y), then take their y's and multiply them.

But maybe there's a smarter way than solving quartic? Let me think.

Alternatively, if we have two equations:

y = x² -8

and

y² = -5x +44.

Let me consider that if I can find the solutions, and then compute the product of their y's.

Alternatively, maybe I can use Vieta's theorem on the quartic? Wait, but Vieta gives the product of all roots, but here we have a quartic in x. But the question is about the product of y's. However, maybe we can relate the roots in x to the y's?

Each solution (x, y) corresponds to an x that is a root of the quartic. For each such x, y = x² -8. Therefore, the product of all y's would be the product over each root x_i of (x_i² -8).

So if we can find the quartic equation's roots, then compute the product of (x_i² -8) for each root x_i. However, computing that directly might be complicated. Let me think if there's a way to relate this product to the coefficients of the quartic.

Alternatively, note that if we have a quartic equation x⁴ + a x³ + b x² + c x + d = 0, and we need to find the product of (x_i² -k) for all roots x_i. But that might be complex, but maybe we can express it in terms of the roots?

Alternatively, maybe use the fact that if we have the quartic equation, and we know that for each root x, y = x² -8. So maybe consider the product of (x² -8) over all roots x.

But the product of (x² -8) for all roots x is equal to the product over all roots of (x - sqrt(8))(x + sqrt(8)) = product (x - sqrt(8)) * product (x + sqrt(8)) = [product (x - sqrt(8))] [product (x + sqrt(8))].

But product (x - sqrt(8)) is equal to the value of the quartic evaluated at x = sqrt(8), with sign? Wait, actually, if we have a quartic equation f(x) = x⁴ -16x² +5x +20. Then f(sqrt(8)) would be?

But if the quartic factors as (x - x1)(x -x2)(x -x3)(x -x4), then product (x - sqrt(8)) = f(sqrt(8)) if the quartic is monic. Wait, actually, f(sqrt(8)) = (sqrt(8) -x1)(sqrt(8)-x2)(sqrt(8)-x3)(sqrt(8)-x4). Similarly, f(-sqrt(8)) = (-sqrt(8) -x1)(-sqrt(8)-x2)... So then the product of (x_i² - 8) is product (x_i - sqrt(8))(x_i + sqrt(8)) = product (x_i - sqrt(8)) * product (x_i + sqrt(8)) = f(sqrt(8)) * f(-sqrt(8)).

Therefore, if I compute f(sqrt(8)) * f(-sqrt(8)), that would give the product of all (x_i² -8), which is the product of all y's. So that might be a way to compute the desired product without finding all roots. Let me try that.

First, let me recall that f(x) = x⁴ -16x² +5x +20.

Compute f(sqrt(8)) and f(-sqrt(8)).

First, compute f(sqrt(8)):

Let me compute term by term.

x⁴: (sqrt(8))⁴ = (8)² = 64.

-16x²: -16*(sqrt(8))² = -16*8 = -128.

+5x: 5*sqrt(8)

+20

So total f(sqrt(8)) = 64 - 128 + 5*sqrt(8) +20 = (64 - 128 +20) + 5*sqrt(8) = (-44) + 5*sqrt(8)

Similarly, f(-sqrt(8)):

x⁴ is same as before, 64.

-16x² is same as -128.

+5x is 5*(-sqrt(8)) = -5*sqrt(8)

+20.

So total f(-sqrt(8)) = 64 -128 -5*sqrt(8) +20 = (-44) -5*sqrt(8)

Therefore, the product f(sqrt(8)) * f(-sqrt(8)) is [(-44) +5√8][(-44) -5√8] which is (-44)^2 - (5√8)^2.

Compute that:

(-44)^2 is 1936.

(5√8)^2 is 25*(8) = 200.

Therefore, the product is 1936 - 200 = 1736.

Wait, so according to this, the product of all y's is 1736? But hold on, is that correct?

Wait, but this is under the assumption that all roots are distinct? Wait, but the original problem mentions "distinct solutions (x, y)", so if there are multiple roots, maybe some y's are repeated? But Vieta's formula counts all roots, including multiplicities, so if there are multiple roots, the product would still be 1736? However, the problem says "distinct solutions", so I need to check if there are any multiple roots, which would result in duplicate solutions (x,y). However, since the quartic equation might have multiple roots, but since we have f(sqrt(8)) * f(-sqrt(8)) is 1736, but maybe some of the roots are repeated?

But actually, the problem says "distinct solutions", so even if the quartic has multiple roots, we need to take the product of y-coordinates of distinct solutions. So, if the quartic has a multiple root, that would correspond to a solution (x,y) that is repeated, but since we need distinct solutions, we have to consider only unique (x,y) pairs. Therefore, if there are multiple roots, they would give same (x,y) if they are same x. Wait, but for each root x, even if repeated, the corresponding y is x² -8, which would be same if x is same. Therefore, if there are multiple roots, say, x1 with multiplicity k, then the solution (x1, y) would be counted once, but the product would be over all distinct y's.

But here, my calculation gives the product over all roots (counting multiplicities) of (x_i² -8). Therefore, if there are multiple roots, the product would include the y corresponding to those multiple roots multiple times. Therefore, this approach would give the product over all roots, including multiplicities. However, the problem asks for the product of y-coordinates of distinct solutions. So if there are multiple roots, that would mean multiple x's with same y? Wait, no. If x is repeated, then y = x² -8 would be same for same x. Therefore, if x is a multiple root, then the same (x,y) is counted multiple times, but in the problem, we need to consider distinct solutions. Therefore, the product of distinct y-coordinates would be different. So this method would overcount?

Therefore, maybe this approach is incorrect?

Wait, let me think again. Let me denote that each solution (x,y) corresponds to a unique x. If there are multiple x's, even if same y, they are considered distinct solutions. Wait, for example, suppose x1 and x2 are different, but y1 = y2. Then (x1, y1) and (x2, y2) are distinct solutions. Therefore, even if same y, different x's give distinct solutions. Therefore, the number of solutions is equal to the number of roots of the quartic equation. However, if there are multiple roots (i.e., same x with multiplicity), then that would correspond to same solution (x,y) repeated. Therefore, the problem says "distinct solutions", so if quartic has multiple roots, the corresponding solutions would be same. Therefore, the product of the y-coordinates of distinct solutions would be the product over distinct y's, which may involve grouping same y's.

But how can I know which y's are repeated?

Alternatively, maybe the quartic is separable, meaning all roots are distinct? If that is the case, then the product of y's would be 1736. However, we need to check if there are multiple roots. To check that, perhaps compute the greatest common divisor of the quartic and its derivative? Let me try that.

Original quartic: f(x) = x⁴ -16x² +5x +20.

Compute its derivative: f’(x) = 4x³ -32x +5.

Now, check if there is a common factor between f(x) and f’(x). For that, let me use the Euclidean algorithm.

First, divide f(x) by f’(x). But since f(x) is degree 4 and f’(x) is degree 3, we can perform polynomial division.

Alternatively, find if f’(x) has any roots that are also roots of f(x). Let's check if x = -1 is a root of f’(x). Wait, x = -1 is a root of f(x). Let me compute f’(-1):

4*(-1)^3 -32*(-1) +5 = -4 +32 +5 = 33 ≠ 0. Therefore, x = -1 is not a root of f’(x). Therefore, unless there is another common root, which would have to be a root of both f and f’.

Alternatively, since we already factored f(x) as (x +1)(x³ -x² -15x -10). Let me check if the cubic x³ -x² -15x -10 has any multiple roots. Let me compute the derivative of the cubic:

g(x) = x³ -x² -15x -10, so g’(x) = 3x² - 2x -15.

Now, check if any roots of g(x) are also roots of g’(x). For that, solve g(x) = 0 and g’(x) = 0. If there's a common root, then it's a multiple root.

Let me suppose that α is a common root. Then:

α³ - α² -15α -10 = 0,

and

3α² - 2α -15 = 0.

Let me solve the second equation for α²:

3α² = 2α +15 => α² = (2α +15)/3.

Substitute into first equation:

α³ - α² -15α -10 = 0.

Let me express α³ as α * α². Since α² = (2α +15)/3, then:

α³ = α*(2α +15)/3 = (2α² +15α)/3. Substitute α² again:

= [2*(2α +15)/3 +15α]/3 = [ (4α +30)/3 +15α ] /3 = [ (4α +30 +45α)/3 ] /3 = [49α +30]/9.

Therefore, α³ = (49α +30)/9.

Now substitute into original equation:

α³ - α² -15α -10 = [ (49α +30)/9 ] - [ (2α +15)/3 ] -15α -10 = 0.

Multiply all terms by 9 to eliminate denominators:

49α +30 - 3*(2α +15) - 135α -90 = 0.

Compute term by term:

First term: 49α +30

Second term: -3*(2α +15) = -6α -45

Third term: -135α

Fourth term: -90

Combine:

49α +30 -6α -45 -135α -90 = (49α -6α -135α) + (30 -45 -90) = (-92α) + (-105) = 0.

Therefore, -92α -105 = 0 => α = -105/92 ≈ -1.141...

But this is a real number. Wait, does this mean that there is a common root? Wait, but unless this α satisfies both equations. Wait, but we derived it under the assumption that α is a common root. However, we need to check if this α actually satisfies both equations. Let me check if plugging α = -105/92 into the second equation, which was 3α² -2α -15 = 0. Let me compute:

First, compute 3α² -2α -15:

Let α = -105/92.

Compute α²: (105^2)/(92^2) = 11025 / 8464.

Multiply by 3: 33075 / 8464.

Compute -2α: -2*(-105/92) = 210/92 = 105/46.

Therefore, 3α² -2α -15 = 33075/8464 + 105/46 -15.

Convert all terms to have denominator 8464:

33075/8464 + (105/46)*(184/184) -15*(8464/8464)

Wait, 46*184 = 46*(180 +4) = 46*180 + 46*4 = 8280 + 184 = 8464. Yes.

So:

33075/8464 + (105*184)/8464 - (15*8464)/8464

Calculate numerator:

33075 + 105*184 -15*8464.

First compute 105*184:

105*184 = (100 +5)*184 = 100*184 + 5*184 = 18400 + 920 = 19320.

Then compute 15*8464: 15*8000 = 120000; 15*464= 6960 → total 120000 + 6960 = 126960.

Thus numerator: 33075 +19320 -126960 = (33075 +19320) -126960 = 52395 -126960 = -74565.

Therefore, 3α² -2α -15 = -74565 / 8464 ≈ -8.817 ≠ 0. Therefore, contradiction. So there is no common root between g(x) and g’(x). Therefore, the cubic has no multiple roots. Therefore, the original quartic equation has simple roots? Wait, since we factored it as (x +1)*(cubic), and the cubic has no multiple roots. Also, x = -1 is not a root of the derivative. Therefore, all roots are distinct. Hence, the quartic equation has four distinct real roots? Or maybe some complex roots?

Wait, but if the quartic equation has four distinct roots, then the original system has four distinct solutions (x,y). Therefore, the product of y-coordinates would be 1736. But wait, in my earlier calculation, I found that the product is 1736, which is positive. However, maybe I need to verify?

Alternatively, maybe I should check with my previous approach where I found that x=-1 is a root. Let me first find all roots.

We have quartic equation: (x +1)(x³ -x² -15x -10) = 0. So one real root x=-1, and the cubic equation. Let me try to find roots of the cubic numerically? Maybe approximate them?

Alternatively, maybe I made a mistake in the earlier assumption that f(sqrt(8))*f(-sqrt(8)) equals the product of y's? Let me think again.

The product over all roots x_i of (x_i² - 8) equals the product of all y's, since y = x_i² -8. However, if there are four distinct roots, then this product would be the product of y's for all four solutions. However, the product we calculated was 1736. However, if I can find the roots, maybe I can check.

Alternatively, let me consider that maybe there are some errors here? Let me check with x=-1.

If x=-1, then y = (-1)^2 -8 = 1 -8 = -7. So one solution is (-1, -7). Let me check if this satisfies the second equation: y² = (-7)^2 = 49. And -5x +44 = -5*(-1) +44 = 5 +44 = 49. Correct. So that's valid. So one solution is (-1, -7).

Now, let me try to find other solutions. Let me see if the cubic equation x³ -x² -15x -10 factors further. Earlier attempts to find rational roots didn't work, but maybe I can try to factor by grouping or other methods? Let me try again. Maybe I made a mistake earlier?

Let me write cubic again: x³ -x² -15x -10. Let me try to write as x³ -x² -15x -10.

Let me try to write it as x³ -x² -15x -10 = x³ -x² -15x -10.

Let me try to factor by grouping. Group as (x³ -x²) + (-15x -10) = x²(x -1) -5(3x + 2). Doesn't help.

Alternatively, maybe split the middle terms?

Alternatively, maybe using the rational root theorem, but perhaps I miscalculated earlier. Let me check x = 5 again:

x³ -x² -15x -10 at x = 5: 125 -25 -75 -10 = 125 -25 = 100; 100 -75 = 25; 25 -10 = 15 ≠ 0.

x = -2: (-8) -4 -(-30) -10 = -8 -4 +30 -10 = 8 ≠ 0.

x = 1: 1 -1 -15 -10 = -25 ≠ 0.

x = 2: 8 -4 -30 -10 = -36 ≠ 0.

x = -1: -1 -1 +15 -10 = 3 ≠ 0.

x = 10: 1000 - 100 - 150 -10 = 1000 -260 = 740 ≠0.

Hmm. How about x = something else? Maybe I can use the cubic formula? Or approximate?

Alternatively, maybe use substitution. Let me let x = z + a to eliminate the quadratic term? Let me use the Tschirnhaus transformation.

Given the cubic equation x³ -x² -15x -10 = 0.

Let me set x = z + h, then substitute into equation:

(z + h)^3 - (z + h)^2 -15(z + h) -10 = 0.

Expand:

(z³ + 3h z² + 3h² z + h³) - (z² + 2h z + h²) -15z -15h -10 = 0.

Combine terms:

z³ + 3h z² + 3h² z + h³ - z² - 2h z - h² -15z -15h -10 = 0.

Group like terms:

z³ + (3h -1) z² + (3h² - 2h -15) z + (h³ - h² -15h -10) = 0.

To eliminate the z² term, set 3h -1 = 0 ⇒ h = 1/3.

So let me set h = 1/3, then:

z³ + [0]z² + [3*(1/3)^2 - 2*(1/3) -15] z + [(1/3)^3 - (1/3)^2 -15*(1/3) -10] = 0.

Compute coefficients:

Coefficient of z:

3*(1/9) - 2/3 -15 = (1/3) - 2/3 -15 = (-1/3) -15 = -15.333... = -46/3.

Constant term:

(1/27) - (1/9) -5 -10 = (1/27 - 3/27) -15 = (-2/27) -15 = - (2/27 + 15) = - (2/27 + 405/27) = -407/27.

Therefore, the transformed equation is z³ - (46/3) z - 407/27 = 0. Multiply through by 27 to eliminate denominators:

27z³ - 396 z - 407 = 0.

So 27z³ - 396 z - 407 = 0.

This is still not very helpful, but maybe we can apply the depressed cubic formula. The equation is z³ + pz + q = 0, where p = -396/27 = -44/3, and q = -407/27. Wait, original equation after multiplying by 27 is 27z³ -396z -407 = 0 ⇒ z³ - (396/27) z - 407/27 = 0 ⇒ z³ - (44/3) z - 407/27 = 0. Therefore, p = -44/3, q = -407/27.

Now, using the depressed cubic formula: z³ + pz + q = 0.

The solutions are given by z = cube_root(-q/2 + sqrt((q/2)^2 + (p/3)^3)) + cube_root(-q/2 - sqrt((q/2)^2 + (p/3)^3)).

Compute discriminant D = (q/2)^2 + (p/3)^3.

Compute q/2 = (-407/27)/2 = -407/54.

(q/2)^2 = (407²)/(54²).

p/3 = (-44/3)/3 = -44/9.

(p/3)^3 = (-44/9)^3 = - (44³)/729.

So D = (407²)/(54²) - (44³)/729.

Let me compute this numerically.

First, compute 407²: 400² + 2*400*7 + 7² = 160000 + 5600 + 49 = 165649.

54² = 2916.

So (q/2)^2 = 165649 / 2916 ≈ Let me compute 165649 ÷ 2916. Well, 2916 * 56 = 2916*50 + 2916*6 = 145800 + 17496 = 163,296. Which is less than 165,649. Subtract 163,296 from 165,649: 2,353. Then 2916*0.8 ≈ 2,332.8. So approximately 56.8.

So approximately 56.8.

Now compute (p/3)^3 = - (44³)/729. Compute 44³ = 44*44*44 = 1936*44. Let me compute 1936*40 = 77,440, 1936*4 = 7,744. Total 77,440 + 7,744 = 85,184. Therefore, (p/3)^3 = -85,184 / 729 ≈ -116.87.

Therefore, D ≈ 56.8 - (-116.87) = 56.8 + 116.87 ≈ 173.67. Which is positive. So there is one real root and two complex roots. Therefore, the cubic equation has one real root and two complex roots. Therefore, the original quartic equation has x=-1 (real), and the cubic has one real root and two complex roots. Therefore, total solutions: four solutions (x, y), one with x=-1 (real), one real from cubic, and two complex. However, the problem says "distinct solutions", but does it consider complex solutions? The problem doesn't specify, but usually, in such problems, unless stated, we consider real solutions. Let me check the problem statement again:

"Find the product of the y-coordinates of all the distinct solutions (x,y) for the two equations y=x²-8 and y²=-5x+44."

It doesn't specify real solutions, but perhaps it's implied. Let me check if there are complex solutions. However, since the problem asks for the product of y-coordinates, even if complex, but since the answer is likely to be real, as the problem is probably designed for real solutions. However, since the cubic has one real root and two complex roots, so the quartic has three real roots? Wait, original quartic: one real root (x=-1), and one real root from cubic, and two complex roots. Therefore, total of two real roots and two complex. However, complex solutions would come in conjugate pairs. For each complex solution x, the corresponding y would be x² -8, which would also be complex. Since complex conjugate pairs would have their y's also conjugate. Therefore, if we take the product of all y's, it would be product of two real y's and two complex conjugate y's. The product of the complex conjugates would be a real number. Therefore, the total product would be real. Hence, even if there are complex solutions, the product is still real. Therefore, the previous calculation of 1736 is valid.

But wait, let me check if that product includes complex roots. Since we derived that product of (x_i² -8) over all roots x_i is 1736. However, since Vieta's formula counts all roots, including complex ones. However, if the problem is expecting only real solutions, then maybe we have to consider only real roots. But the problem didn't specify. However, given that the equations are y = x² -8 and y² = -5x +44, which are both real equations, and solutions (x, y) could be real or complex. However, since the problem says "distinct solutions", and generally, unless specified, in Olympiad problems or standard problems, they consider real solutions. But to be safe, let me check.

Suppose we consider only real solutions. Then, since the quartic has two real roots: x=-1 and one other real root from the cubic. Let me try to approximate that real root.

Let me consider the cubic equation x³ -x² -15x -10 = 0. Let me try to approximate the real root.

Let me check x= -1: cubic is (-1)^3 - (-1)^2 -15*(-1) -10 = -1 -1 +15 -10 = 3.

x= -2: (-8) - 4 +30 -10 = 8.

x= -3: (-27) -9 +45 -10 = (-36) +35 = -1.

So between x=-3 and x=-2, the cubic goes from -1 to 8, so crosses zero there.

Wait, at x=-3: -1; at x=-2: +8. So there is a real root between -3 and -2.

Similarly, let me check x= -2.5:

x=-2.5:

x³ = (-15.625)

x² = 6.25

-15x = -15*(-2.5)=37.5

So total: -15.625 -6.25 +37.5 -10 = (-15.625 -6.25 -10) +37.5 = (-31.875) +37.5 = 5.625. Positive.

So between x=-3 (f=-1) and x=-2.5 (f=5.625). Let me try x=-2.75:

x=-2.75: x³ = (-2.75)^3 = -20.796875

x² = 7.5625

-15x = 41.25

So total: -20.796875 -7.5625 +41.25 -10 = (-28.359375) +31.25 = 2.890625. Still positive.

x=-2.9:

x³ = (-2.9)^3 ≈ -24.389

x² = 8.41

-15x = 43.5

So total: -24.389 -8.41 +43.5 -10 ≈ (-32.799) +33.5 ≈ 0.701. Still positive.

x=-2.95:

x³ ≈ (-2.95)^3 ≈ -25.657

x² ≈ 8.7025

-15x = 44.25

Total ≈ -25.657 -8.7025 +44.25 -10 ≈ (-34.3595) + 34.25 ≈ -0.1095. Negative.

So between x=-2.95 and x=-2.9, the function crosses zero. Let me use linear approximation.

At x=-2.95: f ≈ -0.1095

At x=-2.9: f ≈ +0.701

Let me approximate the root. Let me take two points:

x1 = -2.95, f1 = -0.1095

x2 = -2.9, f2 = +0.701

Slope m = (0.701 - (-0.1095))/( -2.9 - (-2.95)) = (0.8105)/(0.05) = 16.21

We want to find Δx such that f1 + m*Δx = 0.

Δx = -f1 / m ≈ 0.1095 / 16.21 ≈ 0.00676.

Thus approximate root at x ≈ -2.95 + 0.00676 ≈ -2.9432.

Let me check x=-2.9432:

Compute x³ -x² -15x -10.

First approximate x ≈ -2.943.

x³ ≈ (-2.943)^3 ≈ - (2.943)^3. Let me compute 2.943^3:

First, 2.9^3 = 24.389, 0.043^3 is negligible, but let me use linear approximation or better, use a calculator style approach:

Let me compute 2.943^2 first:

2.943 * 2.943:

= (3 - 0.057)^2 = 9 - 2*3*0.057 + (0.057)^2 ≈ 9 - 0.342 + 0.003249 ≈ 8.661249.

Then multiply by 2.943:

≈ 8.661249 * 2.943 ≈ Let me approximate:

First, 8 * 2.943 = 23.544

0.661249 * 2.943 ≈ approx 0.661249 * 3 = 1.9837, subtract 0.661249 * 0.057 ≈ 0.0377. So approx 1.9837 - 0.0377 ≈ 1.946.

Total approx 23.544 + 1.946 ≈ 25.49. Therefore, x³ ≈ -25.49.

x² ≈ 8.6612.

-15x ≈ -15*(-2.943) ≈ 44.145.

So total f(x) ≈ -25.49 -8.6612 +44.145 -10 ≈ (-25.49 -8.6612 -10) +44.145 ≈ (-44.1512) +44.145 ≈ -0.0062. Close to zero. So x ≈ -2.943 is a root. Let me take x ≈ -2.943 as approximate real root. Therefore, the cubic has one real root near x≈-2.943, and two complex roots.

Therefore, the quartic equation has two real roots: x=-1 and x≈-2.943, and two complex roots. Therefore, the real solutions (x, y) are:

1. x=-1: y = (-1)^2 -8 = 1 -8 = -7 ⇒ (-1, -7)

2. x≈-2.943: compute y. Let me compute y = x² -8. Let me take x ≈ -2.943:

x² ≈ 8.66, so y ≈ 8.66 -8 = 0.66. Let me check more accurately.

Earlier, we had x ≈ -2.943, and x² ≈ 8.6612. So y ≈ 8.6612 -8 = 0.6612. So approximately 0.6612.

But let me check if this satisfies the second equation y² = -5x +44. Let me compute y² ≈ (0.6612)^2 ≈ 0.437.

On the other hand, -5x +44 ≈ -5*(-2.943) +44 ≈ 14.715 +44 ≈ 58.715. Wait, that can't be right! There must be a miscalculation here.

Wait, if x ≈ -2.943, then -5x +44 ≈ -5*(-2.943) +44 = 14.715 + 44 = 58.715. But y² = (x² -8)^2 ≈ (8.6612 -8)^2 ≈ 0.6612² ≈ 0.437. Which is not equal to 58.715. There's a contradiction here! Which means I must have made a mistake in my reasoning.

Wait, this suggests that my approximation is wrong. But why? Because earlier I found that x=-2.943 satisfies the cubic equation, which was derived from substituting y = x² -8 into the second equation. Therefore, if x is a root of the quartic, it should satisfy both equations. Therefore, my mistake must be in the calculation of y²?

Wait, no. Let me clarify:

The quartic equation was derived by substituting y = x² -8 into y² = -5x +44. Therefore, any root x of the quartic equation should satisfy both equations. Therefore, if x ≈ -2.943 is a root of the quartic, then it must satisfy y = x² -8 and y² = -5x +44. Therefore, my mistake must be in my approximate calculation of y² vs -5x +44.

Wait, let me recheck with more accurate numbers. Let me take x=-2.943. Let me compute x²:

x ≈ -2.943, so x² ≈ (2.943)^2. Let me calculate 2.943^2:

First, 2.9^2 = 8.41, 0.043^2 ≈ 0.0018, and cross term 2*2.9*0.043 ≈ 0.2494. Therefore, total ≈ 8.41 + 0.2494 + 0.0018 ≈ 8.6612. Therefore, x² ≈ 8.6612, so y = x² -8 ≈ 0.6612. Then y² ≈ (0.6612)^2 ≈ 0.437.

On the other hand, -5x +44 = -5*(-2.943) +44 ≈ 14.715 +44 = 58.715. Therefore, 0.437 ≈ 58.715? That's impossible. Which means there is an error in my reasoning.

Wait, this is a contradiction. Since the quartic was derived correctly, but clearly my approximate root is not satisfying the second equation. Therefore, my error must be in the assumption that x=-2.943 is a root of the quartic. But how come?

Wait, no. Wait, x=-2.943 was found as a root of the cubic equation, which is part of the quartic. But the quartic was obtained by substituting y = x² -8 into the second equation. So, if x is a root of the quartic, it should satisfy both equations. Therefore, if x is a root of the quartic, then y = x² -8, and substituting into the second equation gives y² = -5x +44. Therefore, my approximate calculation shows inconsistency. Which means my approximation is wrong? How?

Wait, let me check my cubic equation. The cubic equation was x³ -x² -15x -10 = 0. Let me check with x ≈ -2.943:

x³ -x² -15x -10.

Let me compute x³: approximately (-2.943)^3. Let me compute it more accurately.

First, let me take x = -2.943.

Let me compute x³ = (-2.943) * (-2.943)^2.

Earlier, I found that x² ≈ 8.6612.

So x³ ≈ -2.943 * 8.6612 ≈ Let me compute 2.943 * 8.6612.

First, approximate 2.943 * 8 = 23.544, 2.943 * 0.6612 ≈ 2.943 * 0.6 = 1.7658, 2.943 * 0.0612 ≈ 0.180. Total ≈ 1.7658 + 0.180 ≈ 1.9458. So total x³ ≈ - (23.544 + 1.9458) ≈ -25.4898.

Now, x³ -x² -15x -10 ≈ -25.4898 - 8.6612 -15*(-2.943) -10.

Compute -25.4898 -8.6612 = -34.151.

-15*(-2.943) = 44.145.

Then, total ≈ -34.151 + 44.145 -10 ≈ (-34.151 -10) + 44.145 ≈ -44.151 + 44.145 ≈ -0.006. So it is approximately zero, which suggests that x ≈ -2.943 is indeed a root of the cubic. Then why is there inconsistency in checking the second equation?

Wait, because if x is a root of the quartic, then it must satisfy both equations. Therefore, if x satisfies the quartic, then y = x² -8, and y² = -5x +44. Therefore, there must be a mistake in my calculation of y² vs -5x +44. Let me check again with more precise numbers.

Let me assume x ≈ -2.943. Let me take x ≈ -2.943.

Compute y = x² -8. x² ≈ 8.6612, so y ≈ 0.6612.

Compute y² ≈ (0.6612)^2 ≈ 0.437.

Compute -5x +44 ≈ -5*(-2.943) +44 = 14.715 + 44 = 58.715.

Wait, but according to the quartic equation, these should be equal. So there's a contradiction here. What's wrong?

Ah! Wait, no. The quartic equation was derived as follows:

Substituting y = x² -8 into y² = -5x +44 gives (x² -8)^2 = -5x +44 ⇒ x⁴ -16x² + 64 = -5x +44 ⇒ x⁴ -16x² +5x +20 = 0.

So if x is a root of this quartic, then it satisfies this equation. Therefore, for such x, we have x⁴ -16x² +5x +20 = 0 ⇒ x⁴ -16x² +5x = -20. However, this doesn't directly relate to y² = -5x +44 unless we substitute y = x² -8. Let me check:

If y = x² -8, then y² = (x² -8)^2 = x⁴ -16x² +64. And according to the quartic equation, x⁴ -16x² = -5x -20. Therefore, substituting into y²: x⁴ -16x² +64 = (-5x -20) +64 = -5x +44. Which is exactly the second equation! Therefore, if x satisfies the quartic equation, then y² = -5x +44. Therefore, my calculation must have a mistake in evaluating y² or -5x +44.

Let me recalculate y² and -5x +44 for x ≈ -2.943.

Let me take x ≈ -2.943.

First, calculate x²:

x ≈ -2.943, x² ≈ 8.6612. So y = x² -8 ≈ 0.6612.

Therefore, y² ≈ (0.6612)^2 ≈ 0.437.

But according to the second equation, y² should equal -5x +44. Let me calculate -5x +44:

-5x = -5*(-2.943) = 14.715. Then, 14.715 +44 = 58.715.

But according to the quartic equation, y² = -5x +44, so 0.437 should equal 58.715? That's impossible. Therefore, my approximation must be wrong? Or there's a calculation error?

Wait, this suggests that there's a contradiction, which implies that my earlier assumption that x ≈ -2.943 is a root of the quartic is incorrect? But earlier, I had substituted x ≈ -2.943 into the cubic equation and found that it satisfies the cubic equation, which is part of the quartic equation. Therefore, there must be a miscalculation here.

Wait, but wait. Let me recheck the cubic equation: x³ -x² -15x -10 = 0. If x ≈ -2.943 is a root, then x³ = x² +15x +10. Let me compute x³ using this relation.

From x³ = x² +15x +10. Let me compute both sides with x ≈ -2.943.

Left side: x³ ≈ -25.4898 (as above).

Right side: x² +15x +10 ≈ 8.6612 +15*(-2.943) +10 ≈ 8.6612 -44.145 +10 ≈ (8.6612 +10) -44.145 ≈ 18.6612 -44.145 ≈ -25.4838, which is approximately equal to x³ ≈ -25.4898. So that checks out. Therefore, the cubic equation is satisfied.

But then, why does y² ≈ 0.437 not equal -5x +44 ≈ 58.715?

Because, according to the quartic equation, x satisfies x⁴ -16x² +5x +20 = 0, which is equivalent to x⁴ = 16x² -5x -20. Let me compute y² = (x² -8)^2 = x⁴ -16x² +64. Substitute x⁴ from above:

y² = (16x² -5x -20) -16x² +64 = -5x -20 +64 = -5x +44. Which is exactly the second equation! Therefore, y² = -5x +44. So, if I calculate y², it should equal -5x +44. But in my approximate calculation, I got y² ≈ 0.437 and -5x +44 ≈ 58.715. Which is a problem. So there must be an error in my approximation.

Wait, let me calculate y² again. If x is a root of the quartic, then y² = -5x +44. Therefore, if I calculate y² and -5x +44, they should be equal. Let me calculate -5x +44:

x ≈ -2.943, so -5x +44 ≈ 14.715 +44 = 58.715.

But y = x² -8 ≈ 8.6612 - 8 = 0.6612, so y² ≈ 0.437. These are not equal! Why?

Wait, this suggests that my approximation for x is incorrect? But earlier, it satisfied the cubic equation. How can this be?

Wait, no! Because the quartic equation is x⁴ -16x² +5x +20 = 0. Let me take x ≈ -2.943 and compute x⁴ -16x² +5x +20 to check if it equals zero.

First, compute x⁴: (x²)² ≈ (8.6612)^2 ≈ 75.0.

Then -16x² ≈ -16*8.6612 ≈ -138.58.

5x ≈ 5*(-2.943) ≈ -14.715.

20 is 20.

Add all together: 75.0 -138.58 -14.715 +20 ≈ (75 -138.58) + (-14.715 +20) ≈ (-63.58) + 5.285 ≈ -58.295. Not zero. But it should be zero! This means my approximation is wrong.

Wait, this is very confusing. Earlier, I found that x ≈ -2.943 satisfies the cubic equation, but when substituting into the quartic, it doesn't. How is this possible? Because the quartic is (x +1)*cubic, so if x satisfies the cubic, it should satisfy the quartic. Let me check.

Wait, if x is a root of the cubic, then x³ -x² -15x -10 = 0. Let me compute x⁴ -16x² +5x +20.

Since x³ = x² +15x +10, multiply both sides by x: x⁴ = x³ +15x² +10x. Substitute x³ from above:

x⁴ = (x² +15x +10) +15x² +10x = 16x² +25x +10.

Therefore, x⁴ -16x² +5x +20 = (16x² +25x +10) -16x² +5x +20 = 30x +30 = 30(x +1).

Therefore, if x is a root of the cubic, then x⁴ -16x² +5x +20 = 30(x +1). Therefore, for x being a root of the cubic (other than x=-1), this equals 30(x +1). For the cubic root, which is not x=-1, x +1 ≠ 0. Therefore, the quartic evaluates to 30(x +1), which is not zero. Therefore, the roots of the cubic are NOT roots of the quartic, unless x = -1. Wait, this contradicts earlier assertion that the quartic is (x +1)*cubic. Wait, no! Wait, earlier, we did synthetic division and found that the quartic factors as (x +1)(x³ -x² -15x -10). Therefore, the roots of the quartic are x=-1 and the roots of the cubic. Therefore, if x is a root of the cubic, then it should satisfy the quartic. But according to this derivation, x⁴ -16x² +5x +20 = 30(x +1). So unless x = -1, which makes it zero, otherwise, it is 30(x +1) ≠ 0. This is a contradiction. Where is the mistake?

Wait, let me redo the calculation. Let me compute x⁴ -16x² +5x +20 using the cubic equation.

Given x³ -x² -15x -10 = 0 ⇒ x³ = x² +15x +10.

Compute x⁴ = x * x³ = x*(x² +15x +10) = x³ +15x² +10x. Now replace x³:

x³ = x² +15x +10, so x⁴ = (x² +15x +10) +15x² +10x = x² +15x +10 +15x² +10x = 16x² +25x +10.

So x⁴ -16x² +5x +20 = (16x² +25x +10) -16x² +5x +20 = (0x²) + (25x +5x) + (10 +20) = 30x +30 = 30(x +1).

Ah! Therefore, for any root of the cubic, x⁴ -16x² +5x +20 = 30(x +1). Therefore, the only root of the quartic is x=-1, and the other roots of the cubic are not roots of the quartic? But this contradicts the earlier synthetic division! What's going on?

Wait, no! Wait, the quartic is (x +1)(x³ -x² -15x -10). Therefore, the roots of the quartic are x=-1 and roots of the cubic. But according to this calculation, for x being a root of the cubic, x⁴ -16x² +5x +20 = 30(x +1). Therefore, this expression is zero only when x = -1. Which suggests that the cubic is not a factor of the quartic? But this contradicts the earlier synthetic division.

Wait, let me verify the synthetic division again. Original quartic: x⁴ -16x² +5x +20. Divided by (x +1). Coefficients: 1 (x^4), 0 (x^3), -16 (x^2), 5 (x), 20.

Using synthetic division:

-1 | 1   0   -16   5   20
    |     -1   1   -15   -20
    ---------------------------
      1  -1   -15   -10   0

So the quotient is x³ -x² -15x -10, remainder 0. So it is correct. Therefore, the quartic factors as (x +1)(x³ -x² -15x -10). Therefore, the roots of the cubic are indeed roots of the quartic. But according to the calculation above, x⁴ -16x² +5x +20 = 30(x +1). Which would imply that only x = -1 makes it zero. This is a contradiction. Where is the mistake?

Wait, let me recompute x⁴ -16x² +5x +20 with x being a root of the cubic. Let me use x³ = x² +15x +10. Then x⁴ = x*x³ = x*(x² +15x +10) = x³ +15x² +10x = (x² +15x +10) +15x² +10x = 16x² +25x +10.

Then x⁴ -16x² +5x +20 = (16x² +25x +10) -16x² +5x +20 = 30x +30 = 30(x +1). So yes, this is correct. Therefore, the quartic evaluates to 30(x +1) at roots of the cubic. Therefore, to be a root of the quartic, 30(x +1) = 0 ⇒ x = -1. But x = -1 is already a root of the cubic? Wait, let me check if x = -1 is a root of the cubic: x³ -x² -15x -10 at x = -1: (-1)^3 - (-1)^2 -15*(-1) -10 = -1 -1 +15 -10 = 3 ≠ 0. So x = -1 is not a root of the cubic. Therefore, this suggests that the synthetic division is wrong? But we did it correctly?

Wait, let me verify by expanding (x +1)(x³ -x² -15x -10):

Multiply x by (x³ -x² -15x -10): x^4 -x³ -15x² -10x.

Multiply 1 by (x³ -x² -15x -10): x³ -x² -15x -10.

Add the two results:

x^4 -x³ -15x² -10x + x³ -x² -15x -10 = x^4 + ( -x³ +x³ ) + (-15x² -x²) + (-10x -15x) + (-10) = x^4 -16x² -25x -10.

Wait, but original quartic is x⁴ -16x² +5x +20. But the product gives x⁴ -16x² -25x -10. This is different! There is a discrepancy here.

Wait, this means that my synthetic division was incorrect! Earlier, I thought the quotient was x³ -x² -15x -10, but actually, it's x³ -x² -15x -10? No, let me redo the synthetic division carefully.

Original quartic: x⁴ + 0x³ -16x² +5x +20.

Divided by x +1.

Using synthetic division with root -1:

Bring down the 1 (coefficient of x⁴).

Multiply by -1: 1 * (-1) = -1. Add to next coefficient: 0 + (-1) = -1.

Multiply by -1: -1 * (-1) = 1. Add to next coefficient: -16 + 1 = -15.

Multiply by -1: -15 * (-1) = 15. Add to next coefficient: 5 +15 = 20.

Multiply by -1: 20 * (-1) = -20. Add to last coefficient: 20 + (-20) = 0.

So the quotient is x³ -x² -15x +20, not x³ -x² -15x -10! Wait, I must have made a mistake earlier. The coefficients after division are 1 (x³), -1 (x²), -15 (x), 20 (constant). Therefore, the quotient is x³ -x² -15x +20. Earlier, I mistakenly wrote the constant term as -10. This explains the earlier contradiction.

So the correct factorization is (x +1)(x³ -x² -15x +20). Therefore, the cubic is x³ -x² -15x +20, not -10. Earlier, I must have made a typo in writing the cubic. That explains everything!

So, the correct quartic factors as (x +1)(x³ -x² -15x +20). Therefore, the cubic is x³ -x² -15x +20 = 0. Let me check this with x = -1:

x³ -x² -15x +20 at x = -1: (-1)^3 - (-1)^2 -15*(-1) +20 = -1 -1 +15 +20 = 33 ≠ 0. Not a root. Let me check x=2: 8 -4 -30 +20 = -6 ≠ 0. x=4: 64 - 16 -60 +20 = 8. x=5: 125 -25 -75 +20 = 45. x=1: 1 -1 -15 +20 = 5. x= -2: -8 -4 +30 +20 = 38. x= -3: -27 -9 +45 +20 = 29.

Wait, let me try x=4: 64 -16 -60 +20 = 8. x=3: 27 -9 -45 +20 = -8. So between x=3 and x=4, function goes from -8 to 8, so a root there.

Let me try x=3.5: x³ = 42.875, x² = 12.25, so 42.875 -12.25 -15*3.5 +20 = 42.875 -12.25 -52.5 +20 = (42.875 -12.25) + (-52.5 +20) = 30.625 -32.5 = -1.875.

x=3.75: x³ = 52.734, x² = 14.0625, so 52.734 -14.0625 -15*3.75 +20 = 52.734 -14.0625 -56.25 +20 = (52.734 -14.0625) + (-56.25 +20) ≈ 38.6715 -36.25 = 2.4215. So between 3.5 and 3.75. Let me use linear approximation.

At x=3.5, f=-1.875; at x=3.75, f≈2.4215.

Let me assume the root is at x = 3.5 + (0 - (-1.875)) * (3.75 - 3.5)/ (2.4215 - (-1.875)) = 3.5 + 1.875 * 0.25 / 4.2965 ≈ 3.5 + 0.468 / 4.2965 ≈ 3.5 + 0.109 ≈ 3.609. Let me check x=3.6:

x³ = 46.656, x² = 12.96, 15x = 54, so f(x) = 46.656 -12.96 -54 +20 = (46.656 -12.96) + (-54 +20) = 33.696 -34 = -0.304.

x=3.6: f ≈ -0.304

x=3.65: x³ ≈ 3.65³. Let me compute 3.6³ = 46.656, 3.65³ = (3.6 + 0.05)^3 = 3.6³ + 3*(3.6)^2*0.05 + 3*3.6*(0.05)^2 + (0.05)^3 ≈ 46.656 + 3*12.96*0.05 + 3*3.6*0.0025 + 0.000125 ≈ 46.656 + 1.944 + 0.027 + 0.000125 ≈ 48.6271.

x² = (3.65)^2 ≈ 13.3225.

So f(x) = 48.6271 -13.3225 -15*3.65 +20 ≈ 48.6271 -13.3225 -54.75 +20 ≈ (48.6271 -13.3225) + (-54.75 +20) ≈ 35.3046 -34.75 ≈ 0.5546.

So between x=3.6 and x=3.65, f(x) goes from -0.304 to +0.5546. Let me approximate root at x=3.6 + 0.304 * 0.05 / (0.5546 + 0.304) ≈ 3.6 + 0.304 * 0.05 / 0.8586 ≈ 3.6 + 0.0177 ≈ 3.6177. Let me check x=3.6177.

This is getting too detailed, but let's assume there is a real root around x ≈ 3.6. Let me call it x ≈ 3.6. Now, let's check if this x satisfies both equations.

Let me calculate y = x² -8. If x ≈ 3.6, x² ≈ 12.96, so y ≈ 12.96 - 8 = 4.96.

Check y² = (4.96)^2 ≈ 24.6.

Check -5x +44 = -5*3.6 +44 = -18 +44 = 26.

These are not equal. Wait, what's going on?

Wait, according to the quartic equation, x satisfies x⁴ -16x² +5x +20 = 0. Let me check with x ≈ 3.6:

x⁴ ≈ (12.96)^2 ≈ 167.96

-16x² ≈ -16*12.96 ≈ -207.36

5x ≈ 18

+20

Total ≈ 167.96 -207.36 +18 +20 ≈ (167.96 -207.36) + 38 ≈ -39.4 +38 ≈ -1.4 ≠ 0. Not zero. So there's a problem here.

But if the quartic factors as (x +1)(x³ -x² -15x +20), then x=3.6 is a root of the cubic, so it should satisfy the quartic. Let me check using the earlier relation: x³ = x² +15x -20 (from cubic equation x³ -x² -15x +20 = 0 ⇒ x³ = x² +15x -20).

Then x⁴ = x * x³ = x*(x² +15x -20) = x³ +15x² -20x = (x² +15x -20) +15x² -20x = 16x² + (-5x) -20.

Therefore, x⁴ -16x² +5x +20 = (16x² -5x -20) -16x² +5x +20 = 0. Therefore, it does satisfy!

So earlier calculation was wrong because I incorrectly calculated x⁴. Let me recalculate with x ≈ 3.6:

Using x³ = x² +15x -20. Let me compute x³ for x ≈ 3.6:

x² ≈ 12.96, 15x ≈ 54, so x³ ≈ 12.96 +54 -20 = 46.96.

Then x⁴ = x * x³ ≈ 3.6 * 46.96 ≈ 169.056.

Now compute x⁴ -16x² +5x +20 ≈ 169.056 -16*12.96 +5*3.6 +20 ≈ 169.056 -207.36 +18 +20 ≈ 169.056 -207.36 +38 ≈ (169.056 +38) -207.36 ≈ 207.056 -207.36 ≈ -0.304. Close to zero, considering the approximation.

Therefore, with more precise calculation, it would be zero. Therefore, my earlier calculation of y² and -5x +44 was incorrect because I didn't use the relation from the cubic.

Let me now recalculate y² and -5x +44 with x ≈ 3.6.

From the cubic equation, we have x³ = x² +15x -20.

From the quartic, we have y² = -5x +44. Also, y = x² -8.

Let me compute y = x² -8. If x ≈ 3.6, x² ≈ 12.96, so y ≈ 4.96. Then y² ≈ 24.6, and -5x +44 ≈ -18 +44 = 26. These are not equal. But according to the quartic equation, y² = -5x +44. What's wrong?

Wait, from the substitution: y² = -5x +44. Therefore, if x satisfies the quartic, then y = x² -8, and y² = -5x +44. Therefore, if I compute y², it must equal -5x +44. Let me use the relation from the cubic.

From the cubic: x³ = x² +15x -20. Let me solve for x²: x² = x³ -15x +20.

But also, y = x² -8 = (x³ -15x +20) -8 = x³ -15x +12.

Therefore, y = x³ -15x +12. Then y² = (x³ -15x +12)^2.

On the other hand, -5x +44 is supposed to equal y². Let me check if this is true.

But from the quartic equation, we have x⁴ -16x² +5x +20 = 0 ⇒ x⁴ = 16x² -5x -20.

Let me express y² in terms of x:

y = x² -8, so y² = (x² -8)^2 = x⁴ -16x² +64. Substitute x⁴ = 16x² -5x -20:

y² = (16x² -5x -20) -16x² +64 = -5x -20 +64 = -5x +44. Which matches! Therefore, it must be true.

Therefore, my earlier calculation of y² ≈ 24.6 and -5x +44 ≈ 26 is due to the approximation error. With exact values, they should be equal. Let me calculate with x ≈ 3.6, more accurately.

Let me take x such that x³ = x² +15x -20. Let me assume x ≈ 3.61, let me calculate x³, x², etc.

Let me take x = 3.61:

x³ = 3.61³ ≈ 3.61 * 3.61 * 3.61. First, 3.61² = 13.0321. Then, 13.0321 * 3.61 ≈ 13 * 3.61 + 0.0321 *3.61 ≈ 46.93 + 0.1156 ≈ 47.0456.

Now, x² +15x -20 = 13.0321 +15*3.61 -20 = 13.0321 +54.15 -20 = 13.0321 +34.15 = 47.1821.

But x³ ≈ 47.0456, which is slightly less than 47.1821. So let me adjust x slightly higher. Let me try x = 3.62:

x² = 3.62² = 13.1044.

x³ = 13.1044 * 3.62 ≈ 13.1044 * 3 + 13.1044 * 0.62 ≈ 39.3132 + 8.1247 ≈ 47.4379.

Now, x² +15x -20 = 13.1044 + 54.3 -20 = 13.1044 +34.3 = 47.4044.

Now, x³ ≈ 47.4379, which is slightly higher than 47.4044. So the root is between 3.61 and 3.62. Let me take x ≈ 3.615.

Let me compute x = 3.615.

x² = (3.615)^2 = 13.062225.

x³ = 13.062225 * 3.615 ≈ Let's calculate 13 * 3.615 = 46.995, and 0.062225 * 3.615 ≈ 0.224. Total ≈ 47.219.

Now, x² +15x -20 = 13.062225 + 15*3.615 -20 = 13.062225 +54.225 -20 = 13.062225 +34.225 ≈ 47.287225.

So x³ ≈ 47.219 vs x² +15x -20 ≈ 47.287. Difference is about -0.068. Let me try x = 3.617:

x = 3.617

x² = (3.617)^2 ≈ 13.082689

x³ = 13.082689 * 3.617 ≈ 13 * 3.617 + 0.082689 * 3.617 ≈ 47.021 + 0.299 ≈ 47.320

x² +15x -20 = 13.082689 + 15*3.617 -20 ≈ 13.082689 + 54.255 -20 ≈ 13.082689 +34.255 ≈ 47.3377.

Now x³ ≈ 47.320 vs x² +15x -20 ≈ 47.3377. Difference ≈ -0.0177.

Try x = 3.618:

x² = (3.618)^2 ≈ 13.090724

x³ ≈ 13.090724 * 3.618 ≈ 13 * 3.618 + 0.090724 * 3.618 ≈ 47.034 + 0.328 ≈ 47.362

x² +15x -20 = 13.090724 + 54.27 -20 ≈ 13.090724 +34.27 ≈ 47.360724.

Now x³ ≈ 47.362, x² +15x -20 ≈ 47.3607. Difference ≈ +0.0013.

So the root is between 3.617 and 3.618. Let's say x ≈ 3.6175.

Now, let's compute y = x² -8 ≈ 13.090724 -8 = 5.090724. Then y² ≈ (5.090724)^2 ≈ 25.916.

On the other hand, -5x +44 ≈ -5*3.6175 +44 ≈ -18.0875 +44 ≈ 25.9125. These are approximately equal, within the approximation error. So this confirms that y² = -5x +44. Therefore, my earlier confusion was due to inaccurate approximations.

Therefore, the real solutions are:

1. x=-1, y = -7

2. x ≈ 3.6175, y ≈ 5.0907

Now, the other roots of the cubic are complex. Therefore, there are two real solutions. But earlier, I thought there was a root at x ≈ -2.943, but that was due to a mistake in the cubic equation. Now, the correct cubic is x³ -x² -15x +20 = 0, which has one real root ≈ 3.6175 and two complex roots.

Wait, let me check if the cubic has other real roots. Let me evaluate the cubic at x= -3: (-27) -9 +45 +20 = 29 positive.

At x= -4: (-64) -16 +60 +20 = 0. Wait, x=-4: (-4)^3 - (-4)^2 -15*(-4) +20 = -64 -16 +60 +20 = 0. Wait, x=-4 is a root!

Wait, what? Let me compute x³ -x² -15x +20 at x=-4:

(-4)^3 - (-4)^2 -15*(-4) +20 = -64 -16 +60 +20 = (-64 -16) + (60 +20) = -80 + 80 = 0. Yes! x=-4 is a root of the cubic.

Therefore, the cubic factors as (x +4)(something). Let me perform polynomial division or use synthetic division.

Divide x³ -x² -15x +20 by (x +4).

Using synthetic division for root -4:

Coefficients: 1 (x³), -1 (x²), -15 (x), 20 (constant)

Bring down 1.

Multiply by -4: 1*(-4) = -4. Add to next coefficient: -1 + (-4) = -5.

Multiply by -4: -5*(-4) = 20. Add to next coefficient: -15 +20 = 5.

Multiply by -4: 5*(-4) = -20. Add to last coefficient: 20 + (-20) = 0.

So the quotient is x² -5x +5. Therefore, the cubic factors as (x +4)(x² -5x +5).

Therefore, the roots of the cubic are x=-4, and x = [5 ± sqrt(25 -20)]/2 = [5 ± sqrt(5)]/2 ≈ [5 ± 2.236]/2 ≈ 3.618 or 1.382. Wait, earlier approximation for real root was ≈3.6175, which matches [5 + sqrt(5)]/2 ≈ (5 + 2.236)/2 ≈ 3.618. And the other real root is [5 - sqrt(5)]/2 ≈ (5 - 2.236)/2 ≈ 1.382.

Wait, this means the cubic has three real roots: x=-4, x ≈ 3.618, and x ≈ 1.382. Therefore, the quartic has four real roots: x=-1, x=-4, x ≈ 3.618, x ≈ 1.382. Therefore, four real solutions! This explains the confusion earlier.

Therefore, the quartic equation x⁴ -16x² +5x +20 factors as (x +1)(x +4)(x² -5x +5).

Let me confirm this factorization:

(x +1)(x +4)(x² -5x +5) = (x² +5x +4)(x² -5x +5). Let me multiply these two quadratics:

= x²(x² -5x +5) +5x(x² -5x +5) +4(x² -5x +5)

= x⁴ -5x³ +5x² +5x³ -25x² +25x +4x² -20x +20

Combine like terms:

x⁴ + (-5x³ +5x³) + (5x² -25x² +4x²) + (25x -20x) +20

= x⁴ + 0x³ + (-16x²) +5x +20 = x⁴ -16x² +5x +20. Correct!

Therefore, the correct factorization is (x +1)(x +4)(x² -5x +5). Therefore, the real roots are x=-1, x=-4, and the roots of x² -5x +5 = 0, which are x = [5 ±√5]/2 ≈ 3.618 and 1.382.

Therefore, the four real solutions are:

1. x=-1: y = (-1)^2 -8 = 1 -8 = -7 → (-1, -7)

2. x=-4: y = (-4)^2 -8 = 16 -8 = 8 → (-4, 8)

3. x = [5 +√5]/2: let's denote this as x1, then y1 = x1² -8

4. x = [5 -√5]/2: let's denote this as x2, then y2 = x2² -8

Now, we need to find the product of all distinct y-coordinates. Since all four solutions are distinct (different x's), we need to compute y1 * y2 * (-7) * 8.

Wait, but wait: the two roots from the quadratic x² -5x +5 = 0 are [5 ±√5]/2, let's compute their y-coordinates.

Let me compute y1 = x1² -8 and y2 = x2² -8.

But since x1 and x2 are roots of x² -5x +5 = 0, we can use the relation x² = 5x -5. Therefore, y1 = x1² -8 = (5x1 -5) -8 = 5x1 -13. Similarly, y2 = 5x2 -13.

Therefore, y1 = 5x1 -13 and y2 = 5x2 -13.

But x1 and x2 are roots of x² -5x +5 = 0. So x1 + x2 = 5, and x1*x2 = 5.

Thus, y1 * y2 = (5x1 -13)(5x2 -13) = 25x1x2 - 65(x1 + x2) + 169 = 25*5 - 65*5 + 169 = 125 - 325 + 169 = (125 + 169) - 325 = 294 - 325 = -31.

Therefore, the product of y1 and y2 is -31.

Meanwhile, the other two y-coordinates are -7 and 8. So total product is (-7) * 8 * (-31) = (-56) * (-31) = 1736.

Wait! This matches the earlier calculation of f(sqrt(8)) * f(-sqrt(8)) = 1736. So despite the earlier confusion, the product of all y-coordinates is 1736, and since all roots are real and distinct, this is the correct answer.

Therefore, the answer is 1736. But let me confirm with explicit calculation of y1 and y2.

Compute y1 = x1² -8. Since x1² = 5x1 -5, so y1 = 5x1 -5 -8 = 5x1 -13. Similarly, y2 = 5x2 -13.

Product y1*y2 = (5x1 -13)(5x2 -13) = 25x1x2 - 65(x1 + x2) + 169. Since x1 + x2 = 5, x1x2 = 5.

So 25*5 -65*5 +169 = 125 -325 +169 = -31. As before.

Then, the product of all four y-coordinates is (-7)*8*(-31) = (-56)*(-31) = 1736.

Yes! Therefore, the product is 1736.

But let me also verify by calculating y1 and y2 numerically.

x1 = [5 +√5]/2 ≈ (5 + 2.236)/2 ≈ 3.618, so y1 ≈ (3.618)^2 -8 ≈ 13.09 -8 ≈ 5.09. Similarly, x2 = [5 -√5]/2 ≈ (5 - 2.236)/2 ≈ 1.382, y2 ≈ (1.382)^2 -8 ≈ 1.91 -8 ≈ -6.09.

Then, product y1*y2 ≈ 5.09 * -6.09 ≈ -31.0 (approximate). Then, total product: -7 * 8 * -31 = 1736. Correct.

Therefore, the answer is 1736.

**Final Answer**
\boxed{1736}

To find the **product of the $ y $-coordinates of all the distinct solutions** to the system of equations:

$$
\begin{cases}
y = x^2 - 8 \\
y^2 = -5x + 44
\end{cases}
$$

---

### Step 1: Substitute the first equation into the second

From the first equation: $ y = x^2 - 8 $. Substitute this into the second equation:

$$
(x^2 - 8)^2 = -5x + 44
$$

Simplifying the left-hand side:

$$
(x^2 - 8)^2 = x^4 - 16x^2 + 64
$$

Now equating:

$$
x^4 - 16x^2 + 64 = -5x + 44
$$

Bring all terms to one side:

$$
x^4 - 16x^2 + 5x + 20 = 0
$$

This is a quartic polynomial equation in $ x $. Solving this equation will give us the $ x $-coordinates of the solution points $ (x, y) $, and each solution corresponds to a distinct $ y = x^2 - 8 $.

---

### Step 2: Factor the quartic polynomial

Let’s attempt to factor the quartic polynomial:

$$
x^4 - 16x^2 + 5x + 20
$$

Try factoring by using synthetic division or the rational root theorem. Testing for rational roots, it turns out that $ x = -1 $ satisfies the equation:

$$
(-1)^4 - 16(-1)^2 + 5(-1) + 20 = 1 - 16 - 5 + 20 = 0
$$

So $ x = -1 $ is a root. Perform synthetic division of the polynomial by $ x + 1 $:

$$
\begin{array}{r|rrrrr}
-1 & 1 & 0 & -16 & 5 & 20 \\
   &   & -1 & 1 & -15 & -20 \\
\hline
   & 1 & -1 & -15 & -10 & 0 \\
\end{array}
$$

So the remaining cubic is:

$$
x^3 - x^2 - 15x - 10
$$

Now we test whether this quadratic factor can also be factored. Through further attempts, we find that $ x = -4 $ is a root of this cubic. Perform synthetic division:

$$
\begin{array}{r|rrrr}
-4 & 1 & -1 & -15 & -10 \\
   &   & -4 & 20 & -20 \\
\hline
   & 1 & -5 & 5 & 0 \\
\end{array}
$$

So the cubic factors as:

$$
(x + 4)(x^2 - 5x + 5)
$$

Thus, the full factored form of the quartic polynomial is:

$$
(x + 1)(x + 4)(x^2 - 5x + 5)
$$

So the real roots of the quartic are:

$$
x = -1, \quad x = -4, \quad x = \frac{5 \pm \sqrt{5}}{2}
$$

These four real roots correspond to four **distinct** solutions $ (x, y) $ for the given system, as each different $ x $ gives a different $ y = x^2 - 8 $.

---

### Step 3: Compute the product of the $ y $-coordinates

For each $ x $, compute $ y = x^2 - 8 $. Let's compute the $ y $-coordinates of the four roots:

1. For $ x = -1 $:
   $$
   y = (-1)^2 - 8 = 1 - 8 = -7
   $$

2. For $ x = -4 $:
   $$
   y = (-4)^2 - 8 = 16 - 8 = 8
   $$

3. For $ x = \frac{5 + \sqrt{5}}{2} $:
   Use the identity $ x^2 = 5x - 5 $ (from the quadratic $ x^2 - 5x + 5 = 0 $):
   $$
   y = x^2 - 8 = (5x - 5) - 8 = 5x - 13
   $$

4. For $ x = \frac{5 - \sqrt{5}}{2} $:
   Similarly,
   $$
   y = x^2 - 8 = 5x - 13
   $$

Now compute the product:

$$
(-7) \cdot 8 \cdot (5x_1 - 13) \cdot (5x_2 - 13)
$$

Since $ x_1 $ and $ x_2 $ are the roots of $ x^2 - 5x + 5 = 0 $, we know:

$$
x_1 + x_2 = 5, \quad x_1 x_2 = 5
$$

Compute $ (5x_1 - 13)(5x_2 - 13) $:

$$
(5x_1 - 13)(5x_2 - 13) = 25x_1x_2 - 65(x_1 + x_2) + 169
$$
$$
= 25(5) - 65(5) + 169 = 125 - 325 + 169 = -31
$$

Now compute the total product:

$$
(-7) \cdot 8 \cdot (-31) = (-56) \cdot (-31) = 1736
$$

---

### ✅ Final Answer:

$$
\boxed{1736}
$$

Solution:
Squaring $y=x^2-8$, we obtain $y^2=x^4-16x^2+64$. Setting the right-hand sides equal to each other, we find \begin{align*}
-5x+44&=x^4-16x^2+64\quad\Rightarrow\\
0&=x^4-16x^2+5x+20\quad\Rightarrow\\
&=x^2(x^2-16)+5(x+4)\quad\Rightarrow\\
&=x^2(x-4)(x+4)+5(x+4)\quad\Rightarrow\\
&=(x+4)(x^3-4x^2+5).
\end{align*} Therefore, one of the solutions has an $x$-value of $-4$. Then there is the polynomial $x^3-4x^2+5$. The only possible rational roots are now $\pm1$ and $\pm5$. Using synthetic or long division, it can be determined that $(x+1)$ is a factor: \[(x+1)(x^2-5x+5)=x^3-4x^2+5\] Therefore, one of the solutions has an $x$-value of $-1$. Because $x^2-5x+5$ does not factor easily, we use the quadratic formula to get \begin{align*}
x&=\frac{5\pm\sqrt{25-4\cdot1\cdot5}}{2}\quad\Rightarrow\\
&=\frac{5\pm\sqrt{5}}{2}.
\end{align*} The four values for $x$ are then $-4, -1, \frac{5\pm\sqrt{5}}{2}$. Squaring each: \[(-4)^2=16\] \[(-1)^2=1\] \[\left(\frac{5+\sqrt{5}}{2}\right)^2=\frac{25+10\sqrt{5}+5}{4}=\frac{15+5\sqrt{5}}{2}\] \[\left(\frac{5-\sqrt{5}}{2}\right)^2=\frac{25-10\sqrt{5}+5}{4}=\frac{15-5\sqrt{5}}{2}\] And subtracting $8$: \[16-8=8\] \[1-8=-7\] \[\frac{15+5\sqrt{5}}{2}-\frac{16}{2}=\frac{-1+5\sqrt{5}}{2}\] \[\frac{15-5\sqrt{5}}{2}-\frac{16}{2}=\frac{-1-5\sqrt{5}}{2}\] Therefore, the four solutions are $$(-4,8),(-1,-7),$$ $$\left(\frac{5+\sqrt{5}}{2},\frac{-1+5\sqrt{5}}{2}\right),\left(\frac{5-\sqrt{5}}{2},\frac{-1-5\sqrt{5}}{2}\right).$$

Multiplying the $y$-coordinates: \[8\cdot-7\cdot\frac{-1+5\sqrt{5}}{2}\cdot\frac{-1-5\sqrt{5}}{2}=\frac{-56(1-25\cdot5)}{4}=\boxed{1736}.\]