Explanation:
Okay, so I need to find P(5) where P(x) is a monic cubic polynomial. Let me start by recalling what a monic polynomial is—it's a polynomial where the leading coefficient is 1. Since it's degree 3, it has the form P(x) = x³ + ax² + bx + c, where a, b, c are constants I need to determine. But maybe there's another way to approach this problem using the given information about remainders?

The problem says that when P(x) is divided by (x - 1)(x - 4), the remainder is R(x). Also, when divided by (x - 2)(x - 3), the remainder is 2R(x). Additionally, we know that P(0) = 5. The goal is to find P(5).

First, let me recall the Remainder Factor Theorem. When a polynomial P(x) is divided by a linear divisor (x - k), the remainder is P(k). However, here the divisors are quadratic: (x - 1)(x - 4) and (x - 2)(x - 3). So, the remainder when dividing by a quadratic polynomial will be of degree less than 2, i.e., linear or constant. Therefore, R(x) must be a linear polynomial, say R(x) = mx + n. Then, the remainder when divided by (x - 2)(x - 3) is 2R(x) = 2mx + 2n.

But how can I use this information? Let me think. If I divide P(x) by (x - 1)(x - 4), the remainder is R(x). So, by the division algorithm, we can write:

P(x) = (x - 1)(x - 4)Q(x) + R(x)

where Q(x) is some polynomial. Since P(x) is cubic and (x - 1)(x - 4) is quadratic, Q(x) must be linear. Similarly, when dividing by (x - 2)(x - 3), we have:

P(x) = (x - 2)(x - 3)S(x) + 2R(x)

where S(x) is another linear polynomial.

But since P(x) is cubic, both Q(x) and S(x) are linear. However, maybe I can use the fact that R(x) is the remainder when divided by (x - 1)(x - 4). Therefore, R(x) must satisfy:

P(1) = R(1) and P(4) = R(4)

Similarly, since the remainder when divided by (x - 2)(x - 3) is 2R(x), then:

P(2) = 2R(2) and P(3) = 2R(3)

So, if I can find R(x), then I can find P(1), P(2), P(3), P(4), and then maybe determine P(x) or at least find P(5).

But R(x) is a linear polynomial, so let me denote R(x) = mx + n. Then:

P(1) = m(1) + n = m + n

P(4) = m(4) + n = 4m + n

Similarly, P(2) = 2R(2) = 2(2m + n) = 4m + 2n

P(3) = 2R(3) = 2(3m + n) = 6m + 2n

So, if I can find P(1), P(2), P(3), P(4), then I can set up equations to solve for m and n, and then maybe find the coefficients of P(x). But since P(x) is monic cubic, maybe I can also use P(0) = 5 to find more equations.

But how do I relate P(1), P(2), P(3), P(4) to the coefficients of P(x)? Let me think. Let me write P(x) as x³ + ax² + bx + c. Then:

P(1) = 1 + a + b + c

P(2) = 8 + 4a + 2b + c

P(3) = 27 + 9a + 3b + c

P(4) = 64 + 16a + 4b + c

But we also have expressions for P(1), P(2), P(3), P(4) in terms of m and n:

P(1) = m + n

P(2) = 4m + 2n

P(3) = 6m + 2n

P(4) = 4m + n

So, we can set up equations:

1 + a + b + c = m + n ...(1)

8 + 4a + 2b + c = 4m + 2n ...(2)

27 + 9a + 3b + c = 6m + 2n ...(3)

64 + 16a + 4b + c = 4m + n ...(4)

Additionally, we know that P(0) = c = 5. So, c = 5. That's helpful. Let me substitute c = 5 into equations (1), (2), (3), (4):

Equation (1): 1 + a + b + 5 = m + n => a + b + 6 = m + n ...(1a)

Equation (2): 8 + 4a + 2b + 5 = 4m + 2n => 13 + 4a + 2b = 4m + 2n ...(2a)

Equation (3): 27 + 9a + 3b + 5 = 6m + 2n => 32 + 9a + 3b = 6m + 2n ...(3a)

Equation (4): 64 + 16a + 4b + 5 = 4m + n => 69 + 16a + 4b = 4m + n ...(4a)

So now, we have four equations with variables a, b, m, n. Let me write them again:

(1a): a + b + 6 = m + n

(2a): 4a + 2b + 13 = 4m + 2n

(3a): 9a + 3b + 32 = 6m + 2n

(4a): 16a + 4b + 69 = 4m + n

So, four equations with four variables. Let me try to solve them step by step.

First, perhaps I can express equations (2a), (3a), (4a) in terms of m and n, and then use equation (1a) to relate them.

Alternatively, maybe subtract equations to eliminate variables. Let me try that.

Let me denote equations (2a) and (3a). Let me subtract (2a) from (3a):

(3a) - (2a): (9a + 3b + 32) - (4a + 2b + 13) = (6m + 2n) - (4m + 2n)

Calculating left side: 5a + b + 19

Right side: 2m

So, 5a + b + 19 = 2m ...(5)

Similarly, let me subtract (2a) from (4a). Wait, but (4a) is 16a + 4b + 69 = 4m + n. Let me see if I can find another equation.

Alternatively, maybe express equations (2a) and (4a) in terms of m and n. Let me think.

Alternatively, let me use equation (1a): m + n = a + b + 6. Let me denote this as equation (1a). Let me also note that equations (2a), (3a), (4a) can be written as:

(2a): 4a + 2b + 13 = 4m + 2n

(3a): 9a + 3b + 32 = 6m + 2n

(4a): 16a + 4b + 69 = 4m + n

Let me try to express equations (2a) and (3a) in terms of m + n and m - n or something like that. Let me see.

First, note that equation (2a): 4a + 2b + 13 = 4m + 2n. Let me divide both sides by 2: 2a + b + 6.5 = 2m + n. Hmm, not sure if helpful.

Alternatively, since from equation (1a): m + n = a + b + 6. Let me denote that as equation (1a). Let me also note that equation (2a) can be written as 4a + 2b + 13 = 4m + 2n. Let me write 4m + 2n as 2*(2m + n). But maybe not helpful.

Alternatively, let me consider equations (2a) and (3a). Let me write them again:

(2a): 4a + 2b + 13 = 4m + 2n

(3a): 9a + 3b + 32 = 6m + 2n

Let me subtract (2a) from (3a):

(9a - 4a) + (3b - 2b) + (32 - 13) = (6m - 4m) + (2n - 2n)

Which is 5a + b + 19 = 2m. That's equation (5) as before.

Similarly, let me take equation (4a): 16a + 4b + 69 = 4m + n. Let me try to relate this with equation (1a). From equation (1a): m + n = a + b + 6. Let me solve for n: n = a + b + 6 - m. Then substitute into equation (4a):

16a + 4b + 69 = 4m + (a + b + 6 - m)

Simplify the right side: 4m + a + b + 6 - m = 3m + a + b + 6

So equation (4a) becomes:

16a + 4b + 69 = 3m + a + b + 6

Bring all terms to the left:

16a + 4b + 69 - 3m - a - b - 6 = 0 => 15a + 3b + 63 - 3m = 0 => 15a + 3b - 3m = -63 => divide both sides by 3: 5a + b - m = -21 ...(6)

So equation (6): 5a + b - m = -21

Earlier, equation (5): 5a + b + 19 = 2m

So, let me write equations (5) and (6):

(5): 5a + b + 19 = 2m

(6): 5a + b - m = -21

Let me subtract equation (6) from equation (5):

(5a + b + 19) - (5a + b - m) = 2m - (-21)

Simplify left side: 5a + b + 19 -5a - b + m = m + 19

Right side: 2m + 21

Therefore:

m + 19 = 2m + 21 => 19 - 21 = 2m - m => -2 = m

So m = -2. Then, substitute m = -2 into equation (6):

5a + b - (-2) = -21 => 5a + b + 2 = -21 => 5a + b = -23 ...(7)

Similarly, substitute m = -2 into equation (5):

5a + b + 19 = 2*(-2) => 5a + b + 19 = -4 => 5a + b = -23. Which is same as equation (7). So that's consistent.

So, we have m = -2. Now, let's find n from equation (1a): m + n = a + b + 6. So, n = a + b + 6 - m = a + b + 6 - (-2) = a + b + 8.

So, n = a + b + 8. Let me keep that in mind.

Now, let me use equation (2a): 4a + 2b + 13 = 4m + 2n. Since m = -2, let's substitute:

Left side: 4a + 2b + 13

Right side: 4*(-2) + 2n = -8 + 2n

So:

4a + 2b + 13 = -8 + 2n => 4a + 2b + 21 = 2n => divide both sides by 2: 2a + b + 10.5 = n

But earlier, we had n = a + b + 8. Therefore:

a + b + 8 = 2a + b + 10.5 => subtract a + b from both sides:

8 = a + 10.5 => a = 8 - 10.5 = -2.5

So, a = -2.5, which is -5/2. Let me write that as a = -5/2.

Now, from equation (7): 5a + b = -23. Let me substitute a = -5/2:

5*(-5/2) + b = -23 => -25/2 + b = -23 => b = -23 + 25/2 = (-46/2 + 25/2) = (-21)/2 = -10.5

So, b = -21/2.

So, now we have a = -5/2, b = -21/2, c = 5. Let me check if these values satisfy equation (1a): m + n = a + b + 6. Let me compute m + n. m = -2, and n = a + b + 8. Let me compute n:

n = (-5/2) + (-21/2) + 8 = (-26/2) + 8 = -13 + 8 = -5. Therefore, m + n = -2 + (-5) = -7. On the other hand, a + b + 6 = (-5/2 -21/2) + 6 = (-26/2) + 6 = -13 + 6 = -7. So that checks out.

Now, let me check equation (2a): 4a + 2b + 13 = 4m + 2n. Let me compute left side:

4*(-5/2) + 2*(-21/2) + 13 = (-10) + (-21) + 13 = (-31) + 13 = -18

Right side: 4*(-2) + 2*(-5) = -8 -10 = -18. Okay, that works.

Let me check equation (3a): 9a + 3b + 32 = 6m + 2n. Left side:

9*(-5/2) + 3*(-21/2) + 32 = (-45/2) + (-63/2) + 32 = (-108/2) + 32 = -54 + 32 = -22

Right side: 6*(-2) + 2*(-5) = -12 -10 = -22. Perfect.

Now equation (4a): 16a + 4b + 69 = 4m + n. Let me compute left side:

16*(-5/2) + 4*(-21/2) + 69 = (-40) + (-42) + 69 = (-82) + 69 = -13

Right side: 4*(-2) + (-5) = -8 -5 = -13. Correct. So all equations are satisfied. Great!

So, now we have a = -5/2, b = -21/2, c = 5. Therefore, the polynomial P(x) is:

P(x) = x³ + (-5/2)x² + (-21/2)x + 5

But maybe I can write this with fractions combined:

P(x) = x³ - (5/2)x² - (21/2)x + 5

Alternatively, to make it easier for calculation, maybe multiply through by 2 to eliminate denominators? But since we need to find P(5), perhaps it's easier to just plug in x = 5 into this expression.

But let me check if there's another way. Alternatively, since we have R(x) = mx + n = -2x -5. Wait, m = -2, n = -5. So R(x) = -2x -5. Let me check if that's consistent with the earlier equations.

Yes, since m = -2, n = -5. So R(x) = -2x -5. Then, when divided by (x - 1)(x - 4), the remainder is R(x). Let me check P(1) and P(4):

P(1) = 1 - 5/2 -21/2 + 5 = 1 + 5 - (5/2 + 21/2) = 6 - 26/2 = 6 -13 = -7. Which should be R(1) = -2(1) -5 = -2 -5 = -7. Correct.

Similarly, P(4) = 64 - (5/2)(16) - (21/2)(4) + 5. Let me compute:

First, 64 + 5 = 69. Then, -(5/2)(16) = -40, -(21/2)(4) = -42. So total: 69 -40 -42 = 69 - 82 = -13. Which should be R(4) = -2(4) -5 = -8 -5 = -13. Correct.

Similarly, check P(2) = 8 - (5/2)(4) - (21/2)(2) + 5 = 8 + 5 - (10 + 21) = 13 - 31 = -18. Which should be 2R(2) = 2*(-2*2 -5) = 2*(-4 -5) = 2*(-9) = -18. Correct.

Similarly, P(3) = 27 - (5/2)(9) - (21/2)(3) + 5 = 27 + 5 - (45/2 + 63/2) = 32 - (108/2) = 32 - 54 = -22. Which should be 2R(3) = 2*(-2*3 -5) = 2*(-6 -5) = 2*(-11) = -22. Correct. So all checks out.

Therefore, the polynomial is correct. Now, the question is to find P(5). Let me compute that.

P(5) = (5)^3 - (5/2)(5)^2 - (21/2)(5) + 5

Compute each term:

5³ = 125

(5/2)(5²) = (5/2)(25) = 125/2

(21/2)(5) = 105/2

So:

P(5) = 125 - 125/2 - 105/2 + 5

Combine the fractions:

-125/2 -105/2 = -230/2 = -115

So:

P(5) = 125 - 115 + 5 = 125 - 115 is 10, plus 5 is 15. Wait, 125 - 115 = 10, then 10 + 5 = 15? Wait, no:

Wait, original expression:

125 - 125/2 - 105/2 + 5

Which is 125 + 5 - (125/2 + 105/2) = 130 - (230/2) = 130 - 115 = 15. Yes, that's correct. So P(5) = 15.

But let me check again step by step:

First term: 5³ = 125

Second term: -(5/2)(5²) = -(5/2)(25) = -125/2

Third term: -(21/2)(5) = -105/2

Fourth term: +5

So total:

125 - 125/2 - 105/2 + 5

Convert 125 and 5 to halves:

125 = 250/2, 5 = 10/2

So:

250/2 - 125/2 - 105/2 + 10/2 = (250 - 125 - 105 + 10)/2 = (250 - 125 is 125; 125 - 105 is 20; 20 + 10 is 30)/2 = 30/2 = 15. Yes, same result. So P(5) = 15.

But let me check if there's another way to compute P(5) without going through all that. Maybe using the information about the remainders?

Alternatively, since P(x) = (x - 1)(x - 4)Q(x) + R(x), and also P(x) = (x - 2)(x - 3)S(x) + 2R(x). But since we already found P(x), maybe that's not necessary. But just to confirm, maybe we can use another approach?

Alternatively, since we know R(x) = -2x -5, then P(x) = (x - 1)(x - 4)Q(x) - 2x -5. Since P(x) is cubic, Q(x) is linear. Let me write Q(x) = px + q. Then:

P(x) = (x - 1)(x - 4)(px + q) - 2x -5

But since P(x) is monic cubic, the leading term is x³. Let me check the leading term of (x - 1)(x - 4)(px + q). The leading term is x * x * px = p x³. Since P(x) is monic, p must be 1. Therefore, Q(x) = x + q. Therefore:

P(x) = (x - 1)(x - 4)(x + q) - 2x -5

Let me expand this:

First compute (x - 1)(x - 4) = x² -5x + 4. Then multiply by (x + q):

(x² -5x + 4)(x + q) = x³ + q x² -5x² -5q x + 4x + 4q = x³ + (q -5)x² + (-5q + 4)x + 4q

Then subtract 2x + 5:

P(x) = x³ + (q -5)x² + (-5q + 4)x + 4q - 2x -5 = x³ + (q -5)x² + (-5q + 4 - 2)x + (4q -5) = x³ + (q -5)x² + (-5q + 2)x + (4q -5)

But we already know that P(x) is x³ - (5/2)x² - (21/2)x + 5. Therefore, equate coefficients:

For x³: 1 = 1, okay.

For x²: q -5 = -5/2 => q = -5/2 + 5 = 5/2

For x term: -5q + 2 = -21/2. Let me check with q = 5/2:

-5*(5/2) + 2 = -25/2 + 4/2 = (-25 + 4)/2 = -21/2. Correct.

For constant term: 4q -5 = 4*(5/2) -5 = 10 -5 = 5. Correct. So that's consistent. Therefore, q = 5/2. Therefore, this method also gives us the same coefficients. So P(x) is indeed as found before.

Therefore, P(5) = 15. So the answer is 15. Wait, but let me check once more with another approach.

Alternatively, since we know P(0) = 5, which we used. Also, since we have P(x) expressed in terms of R(x), maybe we can use interpolation?

But since we already have the answer through two methods, and all checks out, I think 15 is correct. However, just to be thorough, let me compute P(5) using the polynomial expression:

P(x) = x³ - (5/2)x² - (21/2)x + 5

So P(5) = 125 - (5/2)(25) - (21/2)(5) + 5

Compute each term:

(5/2)(25) = 125/2 = 62.5

(21/2)(5) = 105/2 = 52.5

So:

125 - 62.5 - 52.5 + 5 = 125 - 62.5 = 62.5; 62.5 - 52.5 = 10; 10 + 5 = 15. Yep, same result.

Therefore, I think the answer is 15. But let me check once more if there's a possibility of miscalculations.

Wait, but let me think again. The problem says that P(x) is monic. Our polynomial is x³ - (5/2)x² - (21/2)x + 5. The leading coefficient is 1, so that's correct. So that's okay.

Alternatively, maybe there's a smarter way using the Chinese Remainder Theorem or something? Let me think.

Since P(x) ≡ R(x) mod (x - 1)(x - 4), and P(x) ≡ 2R(x) mod (x - 2)(x - 3). Also, since (x - 1)(x - 4) and (x - 2)(x - 3) are coprime? Wait, are they? Let me check if they have common factors. The roots of (x - 1)(x - 4) are 1 and 4, and roots of (x - 2)(x - 3) are 2 and 3. So they are distinct, so the two quadratics are coprime. Therefore, by Chinese Remainder Theorem, there exists a unique polynomial modulo (x - 1)(x - 4)(x - 2)(x - 3) that satisfies both congruences. However, since P(x) is a cubic, and (x - 1)(x - 2)(x - 3)(x - 4) is quartic, so P(x) is determined uniquely by its values at four points? Wait, but maybe not necessary here.

Alternatively, since we have P(x) - R(x) is divisible by (x - 1)(x - 4), and P(x) - 2R(x) is divisible by (x - 2)(x - 3). Therefore, P(x) - R(x) = (x - 1)(x - 4)A(x), and P(x) - 2R(x) = (x - 2)(x - 3)B(x), where A(x) and B(x) are linear polynomials. But since P(x) is cubic, A(x) and B(x) are linear. But since we already solved it through another method, maybe not necessary.

Alternatively, since we know R(x) = -2x -5, then P(x) = (x - 1)(x - 4)Q(x) - 2x -5. Since P(x) is monic cubic, Q(x) is linear, as we saw. But since we already found Q(x) = x + 5/2, then P(x) = (x - 1)(x - 4)(x + 5/2) - 2x -5. Let me compute this expression for x = 5:

First compute (5 - 1)(5 - 4)(5 + 5/2) - 2*5 -5 = (4)(1)(15/2) - 10 -5 = 4*(15/2) -15 = 30 -15 = 15. Same result. So that's another way to compute P(5) directly. Therefore, 15 is correct.

Therefore, I think the answer is 15. So I can write that as \boxed{15}.

**Final Answer**
\boxed{15}

To solve the problem, we are given that $ P(x) $ is a **monic cubic polynomial**, meaning its leading coefficient is 1. We are also told:

- When $ P(x) $ is divided by $ (x - 1)(x - 4) $, the remainder is $ R(x) $.
- When $ P(x) $ is divided by $ (x - 2)(x - 3) $, the remainder is $ 2R(x) $.
- $ P(0) = 5 $.

We are to find $ P(5) $.

---

### Step 1: Understand the Structure of the Polynomial

Since $ P(x) $ is a cubic polynomial, we can write it in the form:

$$
P(x) = x^3 + ax^2 + bx + c
$$

We are told that $ P(0) = 5 $, so:

$$
P(0) = c = 5
$$

So, the polynomial becomes:

$$
P(x) = x^3 + ax^2 + bx + 5
$$

---

### Step 2: Use the Remainder Theorem

Let’s denote the remainder $ R(x) $ as a linear polynomial:

$$
R(x) = mx + n
$$

Then, from the problem:

- $ P(1) = R(1) = m + n $
- $ P(4) = R(4) = 4m + n $
- $ P(2) = 2R(2) = 2(2m + n) = 4m + 2n $
- $ P(3) = 2R(3) = 2(3m + n) = 6m + 2n $

Now, compute $ P(1), P(2), P(3), P(4) $ using the general form of $ P(x) $:

$$
\begin{aligned}
P(1) &= 1 + a + b + 5 = a + b + 6 \\
P(2) &= 8 + 4a + 2b + 5 = 4a + 2b + 13 \\
P(3) &= 27 + 9a + 3b + 5 = 9a + 3b + 32 \\
P(4) &= 64 + 16a + 4b + 5 = 16a + 4b + 69
\end{aligned}
$$

Now equate these expressions to the ones in terms of $ m $ and $ n $:

$$
\begin{aligned}
a + b + 6 &= m + n \quad \text{(1)} \\
4a + 2b + 13 &= 4m + 2n \quad \text{(2)} \\
9a + 3b + 32 &= 6m + 2n \quad \text{(3)} \\
16a + 4b + 69 &= 4m + n \quad \text{(4)}
\end{aligned}
$$

---

### Step 3: Solve the System of Equations

From equation (1):
$$
m + n = a + b + 6 \quad \text{(1a)}
$$

From equation (2):
$$
4a + 2b + 13 = 4m + 2n \quad \text{(2a)}
$$

From equation (3):
$$
9a + 3b + 32 = 6m + 2n \quad \text{(3a)}
$$

From equation (4):
$$
16a + 4b + 69 = 4m + n \quad \text{(4a)}
$$

Subtract (2a) from (3a):

$$
(9a + 3b + 32) - (4a + 2b + 13) = (6m + 2n) - (4m + 2n) \\
5a + b + 19 = 2m \quad \text{(5)}
$$

Subtract (2a) from (4a):

$$
(16a + 4b + 69) - (4a + 2b + 13) = (4m + n) - (4m + 2n) \\
12a + 2b + 56 = -n \quad \text{(6)}
$$

Now solve equations (5) and (6) together.

From (5):
$$
5a + b = 2m - 19 \quad \text{(5a)}
$$

From (6):
$$
12a + 2b = -n - 56 \quad \text{(6a)}
$$

Substitute (5a) into (6a):

$$
12a + 2(2m - 19 - 5a) = -n - 56 \\
12a + 4m - 38 - 10a = -n - 56 \\
2a + 4m - 38 = -n - 56 \\
2a + 4m + 18 = -n \quad \text{(7)}
$$

Now use (1a):
$$
m + n = a + b + 6 \Rightarrow n = a + b + 6 - m
$$

Substitute into (7):

$$
2a + 4m + 18 = -(a + b + 6 - m) \\
2a + 4m + 18 = -a - b - 6 + m \\
3a + 3m + b + 24 = 0 \quad \text{(8)}
$$

Now use (5a):
$$
5a + b = 2m - 19 \Rightarrow b = 2m - 19 - 5a
$$

Substitute into (8):

$$
3a + 3m + (2m - 19 - 5a) + 24 = 0 \\
3a + 3m + 2m - 19 - 5a + 24 = 0 \\
-2a + 5m + 5 = 0 \Rightarrow -2a + 5m = -5 \quad \text{(9)}
$$

Now solve (9) and (5a):

From (5a):
$$
5a + b = 2m - 19
$$

From (9):
$$
-2a + 5m = -5 \Rightarrow 5m = 2a - 5 \Rightarrow m = \frac{2a - 5}{5}
$$

Substitute into (5a):

$$
5a + b = 2\left(\frac{2a - 5}{5}\right) - 19 = \frac{4a - 10}{5} - 19 = \frac{4a - 10 - 95}{5} = \frac{4a - 105}{5}
$$

Multiply both sides by 5:

$$
25a + 5b = 4a - 105 \Rightarrow 21a + 5b = -105 \quad \text{(10)}
$$

Now solve (10) and (9) together.

From (9):
$$
5m = 2a - 5 \Rightarrow m = \frac{2a - 5}{5}
$$

From (10):
$$
21a + 5b = -105 \Rightarrow b = \frac{-105 - 21a}{5}
$$

Now use (1a):
$$
m + n = a + b + 6
$$

Substitute $ m $ and $ b $:

$$
\frac{2a - 5}{5} + n = a + \frac{-105 - 21a}{5} + 6
$$

Multiply through by 5:

$$
2a - 5 + 5n = 5a - 105 - 21a + 30 \\
2a - 5 + 5n = -16a - 75 \\
5n = -18a - 70 \Rightarrow n = -\frac{18a + 70}{5}
$$

Now plug into (1a):

$$
\frac{2a - 5}{5} + \left(-\frac{18a + 70}{5}\right) = a + \frac{-105 - 21a}{5} + 6
$$

Multiply through by 5:

$$
2a - 5 - 18a - 70 = 5a - 105 - 21a + 30 \\
-16a - 75 = -16a - 75
$$

So the system is consistent. Now solve for $ a $ and $ b $.

From (9):
$$
-2a + 5m = -5 \Rightarrow m = \frac{2a - 5}{5}
$$

From (5a):
$$
5a + b = 2m - 19 = 2\left(\frac{2a - 5}{5}\right) - 19 = \frac{4a - 10}{5} - 19 = \frac{4a - 10 - 95}{5} = \frac{4a - 105}{5}
$$

Multiply both sides by 5:

$$
25a + 5b = 4a - 105 \Rightarrow 21a + 5b = -105
$$

Solve for $ b $:

$$
b = \frac{-105 - 21a}{5}
$$

Now plug into (1a):

$$
m + n = a + b + 6
$$

Substitute $ m = \frac{2a - 5}{5} $, $ b = \frac{-105 - 21a}{5} $, and solve for $ a $.

Eventually, solving the system gives:

$$
a = -\frac{5}{2}, \quad b = -\frac{21}{2}, \quad c = 5
$$

---

### Step 4: Final Polynomial and Evaluation

So the polynomial is:

$$
P(x) = x^3 - \frac{5}{2}x^2 - \frac{21}{2}x + 5
$$

Now compute $ P(5) $:

$$
P(5) = 5^3 - \frac{5}{2}(5^2) - \frac{21}{2}(5) + 5 = 125 - \frac{125}{2} - \frac{105}{2} + 5
$$

$$
= 125 - \frac{230}{2} + 5 = 125 - 115 + 5 = 15
$$

---

### Final Answer

$$
\boxed{15}
$$

[Assistant]
Explanation:
Okay, so I need to find P(5) where P(x) is a monic cubic polynomial. Let me start by recalling what a monic polynomial is—it's a polynomial where the leading coefficient is 1. Since it's degree 3, it has the form P(x) = x³ + ax² + bx + c, where a, b, c are constants I need to determine. But maybe there's another way to approach this problem using the given information about remainders?

The problem says that when P(x) is divided by (x - 1)(x - 4), the remainder is R(x). Also, when divided by (x - 2)(x - 3), the remainder is 2R(x). Additionally, we know that P(0) = 5. The goal is to find P(5).

First, let me recall the Remainder Factor Theorem. When a polynomial P(x) is divided by a linear divisor (x - k), the remainder is P(k). However, here the divisors are quadratic: (x - 1)(x - 4) and (x - 2)(x - 3). So, the remainder when dividing by a quadratic polynomial will be of degree less than 2, i.e., linear or constant. Therefore, R(x) must be a linear polynomial, say R(x) = mx + n. Then, the remainder when divided by (x - 2)(x - 3) is 2R(x) = 2mx + 2n.

But how can I use this information? Let me think. If I divide P(x) by (x - 1)(x - 4), the remainder is R(x). So, by the division algorithm, we can write:

P(x) = (x - 1)(x - 4)Q(x) + R(x)

where Q(x) is some polynomial. Since P(x) is cubic and (x - 1)(x - 4) is quadratic, Q(x) must be linear. Similarly, when dividing by (x - 2)(x - 3), we have:

P(x) = (x - 2)(x - 3)S(x) + 2R(x)

where S(x) is another linear polynomial.

But since P(x) is cubic, both Q(x) and S(x) are linear. However, maybe I can use the fact that R(x) is the remainder when divided by (x - 1)(x - 4). Therefore, R(x) must satisfy:

P(1) = R(1) and P(4) = R(4)

Similarly, since the remainder when divided by (x - 2)(x - 3) is 2R(x), then:

P(2) = 2R(2) and P(3) = 2R(3)

So, if I can find R(x), then I can find P(1), P(2), P(3), P(4), and then maybe determine P(x) or at least find P(5).

But R(x) is a linear polynomial, so let me denote R(x) = mx + n. Then:

P(1) = m(1) + n = m + n

P(4) = m(4) + n = 4m + n

Similarly, P(2) = 2R(2) = 2(2m + n) = 4m + 2n

P(3) = 2R(3) = 2(3m + n) = 6m + 2n

So, if I can find P(1), P(2), P(3), P(4), then I can set up equations to solve for m and n, and then maybe find the coefficients of P(x). But since P(x) is monic cubic, maybe I can also use P(0) = 5 to find more equations.

But how do I relate P(1), P(2), P(3), P(4) to the coefficients of P(x)? Let me think. Let me write P(x) as x³ + ax² + bx + c. Then:

P(1) = 1 + a + b + c

P(2) = 8 + 4a + 2b + c

P(3) = 27 + 9a + 3b + c

P(4) = 64 + 16a + 4b + c

But we also have expressions for P(1), P(2), P(3), P(4) in terms of m and n:

P(1) = m + n

P(2) = 4m + 2n

P(3) = 6m + 2n

P(4) = 4m + n

So, we can set up equations:

1 + a + b + c = m + n ...(1)

8 + 4a + 2b + c = 4m + 2n ...(2)

27 + 9a + 3b + c = 6m + 2n ...(3)

64 + 16a + 4b + c = 4m + n ...(4)

Additionally, we know that P(0) = c = 5. So, c = 5. That's helpful. Let me substitute c = 5 into equations (1), (2), (3), (4):

Equation (1): 1 + a + b + 5 = m + n => a + b + 6 = m + n ...(1a)

Equation (2): 8 + 4a + 2b + 5 = 4m + 2n => 13 + 4a + 2b = 4m + 2n ...(2a)

Equation (3): 27 + 9a + 3b + 5 = 6m + 2n => 32 + 9a + 3b = 6m + 2n ...(3a)

Equation (4): 64 + 16a + 4b + 5 = 4m + n => 69 + 16a + 4b = 4m + n ...(4a)

So now, we have four equations with variables a, b, m, n. Let me write them again:

(1a): a + b + 6 = m + n

(2a): 4a + 2b + 13 = 4m + 2n

(3a): 9a + 3b + 32 = 6m + 2n

(4a): 16a + 4b + 69 = 4m + n

So, four equations with four variables. Let me try to solve them step by step.

First, perhaps I can express equations (2a), (3a), (4a) in terms of m and n, and then use equation (1a) to relate them.

Alternatively, maybe subtract equations to eliminate variables. Let me try that.

Let me denote equations (2a) and (3a). Let me subtract (2a) from (3a):

(3a) - (2a): (9a + 3b + 32) - (4a + 2b + 13) = (6m + 2n) - (4m + 2n)

Calculating left side: 5a + b + 19

Right side: 2m

So, 5a + b + 19 = 2m ...(5)

Similarly, let me subtract (2a) from (4a). Wait, but (4a) is 16a + 4b + 69 = 4m + n. Let me see if I can find another equation.

Alternatively, maybe express equations (2a) and (4a) in terms of m and n. Let me think.

Alternatively, let me use equation (1a): m + n = a + b + 6. Let me denote this as equation (1a). Let me also note that equations (2a), (3a), (4a) can be written as:

(2a): 4a + 2b + 13 = 4m + 2n

(3a): 9a + 3b + 32 = 6m + 2n

(4a): 16a + 4b + 69 = 4m + n

Let me try to express equations (2a) and (3a) in terms of m + n and m - n or something like that. Let me see.

First, note that equation (2a): 4a + 2b + 13 = 4m + 2n. Let me divide both sides by 2: 2a + b + 6.5 = 2m + n. Hmm, not sure if helpful.

Alternatively, since from equation (1a): m + n = a + b + 6. Let me denote that as equation (1a). Let me also note that equation (2a) can be written as 4a + 2b + 13 = 4m + 2n. Let me write 4m + 2n as 2*(2m + n). But maybe not helpful.

Alternatively, let me consider equations (2a) and (3a). Let me write them again:

(2a): 4a + 2b + 13 = 4m + 2n

(3a): 9a + 3b + 32 = 6m + 2n

Let me subtract (2a) from (3a):

(9a - 4a) + (3b - 2b) + (32 - 13) = (6m - 4m) + (2n - 2n)

Which is 5a + b + 19 = 2m. That's equation (5) as before.

Similarly, let me take equation (4a): 16a + 4b + 69 = 4m + n. Let me try to relate this with equation (1a). From equation (1a): m + n = a + b + 6. Let me solve for n: n = a + b + 6 - m. Then substitute into equation (4a):

16a + 4b + 69 = 4m + (a + b + 6 - m)

Simplify the right side: 4m + a + b + 6 - m = 3m + a + b + 6

So equation (4a) becomes:

16a + 4b + 69 = 3m + a + b + 6

Bring all terms to the left:

16a + 4b + 69 - 3m - a - b - 6 = 0 => 15a + 3b + 63 - 3m = 0 => 15a + 3b - 3m = -63 => divide both sides by 3: 5a + b - m = -21 ...(6)

So equation (6): 5a + b - m = -21

Earlier, equation (5): 5a + b + 19 = 2m

So, let me write equations (5) and (6):

(5): 5a + b + 19 = 2m

(6): 5a + b - m = -21

Let me subtract equation (6) from equation (5):

(5a + b + 19) - (5a + b - m) = 2m - (-21)

Simplify left side: 5a + b + 19 -5a - b + m = m + 19

Right side: 2m + 21

Therefore:

m + 19 = 2m + 21 => 19 - 21 = 2m - m => -2 = m

So m = -2. Then, substitute m = -2 into equation (6):

5a + b - (-2) = -21 => 5a + b + 2 = -21 => 5a + b = -23 ...(7)

Similarly, substitute m = -2 into equation (5):

5a + b + 19 = 2*(-2) => 5a + b + 19 = -4 => 5a + b = -23. Which is same as equation (7). So that's consistent.

So, we have m = -2. Now, let's find n from equation (1a): m + n = a + b + 6. So, n = a + b + 6 - m = a + b + 6 - (-2) = a + b + 8.

So, n = a + b + 8. Let me keep that in mind.

Now, let me use equation (2a): 4a + 2b + 13 = 4m + 2n. Since m = -2, let's substitute:

Left side: 4a + 2b + 13

Right side: 4*(-2) + 2n = -8 + 2n

So:

4a + 2b + 13 = -8 + 2n => 4a + 2b + 21 = 2n => divide both sides by 2: 2a + b + 10.5 = n

But earlier, we had n = a + b + 8. Therefore:

a + b + 8 = 2a + b + 10.5 => subtract a + b from both sides:

8 = a + 10.5 => a = 8 - 10.5 = -2.5

So, a = -2.5, which is -5/2. Let me write that as a = -5/2.

Now, from equation (7): 5a + b = -23. Let me substitute a = -5/2:

5*(-5/2) + b = -23 => -25/2 + b = -23 => b = -23 + 25/2 = (-46/2 + 25/2) = (-21)/2 = -10.5

So, b = -21/2.

So, now we have a = -5/2, b = -21/2, c = 5. Let me check if these values satisfy equation (1a): m + n = a + b + 6. Let me compute m + n. m = -2, and n = a + b + 8. Let me compute n:

n = (-5/2) + (-21/2) + 8 = (-26/2) + 8 = -13 + 8 = -5. Therefore, m + n = -2 + (-5) = -7. On the other hand, a + b + 6 = (-5/2 -21/2) + 6 = (-26/2) + 6 = -13 + 6 = -7. So that checks out.

Now, let me check equation (2a): 4a + 2b + 13 = 4m + 2n. Let me compute left side:

4*(-5/2) + 2*(-21/2) + 13 = (-10) + (-21) + 13 = (-31) + 13 = -18

Right side: 4*(-2) + 2*(-5) = -8 -10 = -18. Okay, that works.

Let me check equation (3a): 9a + 3b + 32 = 6m + 2n. Left side:

9*(-5/2) + 3*(-21/2) + 32 = (-45/2) + (-63/2) + 32 = (-108/2) + 32 = -54 + 32 = -22

Right side: 6*(-2) + 2*(-5) = -12 -10 = -22. Perfect.

Now equation (4a): 16a + 4b + 69 = 4m + n. Let me compute left side:

16*(-5/2) + 4*(-21/2) + 69 = (-40) + (-42) + 69 = (-82) + 69 = -13

Right side: 4*(-2) + (-5) = -8 -5 = -13. Correct. So all equations are satisfied. Great!

So, now we have a = -5/2, b = -21/2, c = 5. Therefore, the polynomial P(x) is:

P(x) = x³ + (-5/2)x² + (-21/2)x + 5

But maybe I can write this with fractions combined:

P(x) = x³ - (5/2)x² - (21/2)x + 5

Alternatively, to make it easier for calculation, maybe multiply through by 2 to eliminate denominators? But since we need to find P(5), perhaps it's easier to just plug in x = 5 into this expression.

But let me check if there's another way. Alternatively, since we have R(x) = mx + n = -2x -5. Wait, m = -2, n = -5. So R(x) = -2x -5. Let me check if that's consistent with the earlier equations.

Yes, since m = -2, n = -5. So R(x) = -2x -5. Then, when divided by (x - 1)(x - 4), the remainder is R(x). Let me check P(1) and P(4):

P(1) = 1 - 5/2 -21/2 + 5 = 1 + 5 - (5/2 + 21/2) = 6 - 26/2 = 6 -13 = -7. Which should be R(1) = -2(1) -5 = -2 -5 = -7. Correct.

Similarly, P(4) = 64 - (5/2)(16) - (21/2)(4) + 5. Let me compute:

First, 64 + 5 = 69. Then, -(5/2)(16) = -40, -(21/2)(4) = -42. So total: 69 -40 -42 = 69 - 82 = -13. Which should be R(4) = -2(4) -5 = -8 -5 = -13. Correct.

Similarly, check P(2) = 8 - (5/2)(4) - (21/2)(2) + 5 = 8 + 5 - (10 + 21) = 13 - 31 = -18. Which should be 2R(2) = 2*(-2*2 -5) = 2*(-4 -5) = 2*(-9) = -18. Correct.

Similarly, P(3) = 27 - (5/2)(9) - (21/2)(3) + 5 = 27 + 5 - (45/2 + 63/2) = 32 - (108/2) = 32 - 54 = -22. Which should be 2R(3) = 2*(-2*3 -5) = 2*(-6 -5) = 2*(-11) = -22. Correct. So all checks out.

Therefore, the polynomial is correct. Now, the question is to find P(5). Let me compute that.

P(5) = (5)^3 - (5/2)(5)^2 - (21/2)(5) + 5

Compute each term:

5³ = 125

(5/2)(5²) = (5/2)(25) = 125/2

(21/2)(5) = 105/2

So:

P(5) = 125 - 125/2 - 105/2 + 5

Combine the fractions:

-125/2 -105/2 = -230/2 = -115

So:

P(5) = 125 - 115 + 5 = 125 - 115 is 10, plus 5 is 15. Wait, 125 - 115 = 10, then 10 + 5 = 15? Wait, no:

Wait, original expression:

125 - 125/2 - 105/2 + 5

Which is 125 + 5 - (125/2 + 105/2) = 130 - (230/2) = 130 - 115 = 15. Yes, that's correct. So P(5) = 15.

But let me check again step by step:

First term: 5³ = 125

Second term: -(5/2)(5²) = -(5/2)(25) = -125/2

Third term: -(21/2)(5) = -105/2

Fourth term: +5

So total:

125 - 125/2 - 105/2 + 5

Convert 125 and 5 to halves:

125 = 250/2, 5 = 10/2

So:

250/2 - 125/2 - 105/2 + 10/2 = (250 - 125 - 105 + 10)/2 = (250 - 125 is 125; 125 - 105 is 20; 20 + 10 is 30)/2 = 30/2 = 15. Yes, same result. So P(5) = 15.

But let me check if there's another way to compute P(5) without going through all that. Maybe using the information about the remainders?

Alternatively, since P(x) = (x - 1)(x - 4)Q(x) + R(x), and also P(x) = (x - 2)(x - 3)S(x) + 2R(x). But since we already found P(x), maybe that's not necessary. But just to confirm, maybe we can use another approach?

Alternatively, since we know R(x) = -2x -5, then P(x) = (x - 1)(x - 4)Q(x) - 2x -5. Since P(x) is cubic, Q(x) is linear. Let me write Q(x) = px + q. Then:

P(x) = (x - 1)(x - 4)(px + q) - 2x -5

But since P(x) is monic cubic, the leading term is x³. Let me check the leading term of (x - 1)(x - 4)(px + q). The leading term is x * x * px = p x³. Since P(x) is monic, p must be 1. Therefore, Q(x) = x + q. Therefore:

P(x) = (x - 1)(x - 4)(x + q) - 2x -5

Let me expand this:

First compute (x - 1)(x - 4) = x² -5x + 4. Then multiply by (x + q):

(x² -5x + 4)(x + q) = x³ + q x² -5x² -5q x + 4x + 4q = x³ + (q -5)x² + (-5q + 4)x + 4q

Then subtract 2x + 5:

P(x) = x³ + (q -5)x² + (-5q + 4)x + 4q - 2x -5 = x³ + (q -5)x² + (-5q + 4 - 2)x + (4q -5) = x³ + (q -5)x² + (-5q + 2)x + (4q -5)

But we already know that P(x) is x³ - (5/2)x² - (21/2)x + 5. Therefore, equate coefficients:

For x³: 1 = 1, okay.

For x²: q -5 = -5/2 => q = -5/2 + 5 = 5/2

For x term: -5q + 2 = -21/2. Let me check with q = 5/2:

-5*(5/2) + 2 = -25/2 + 4/2 = (-25 + 4)/2 = -21/2. Correct.

For constant term: 4q -5 = 4*(5/2) -5 = 10 -5 = 5. Correct. So that's consistent. Therefore, q = 5/2. Therefore, this method also gives us the same coefficients. So P(x) is indeed as found before.

Therefore, P(5) = 15. So the answer is 15. Wait, but let me check once more with another approach.

Alternatively, since we know P(0) = 5, which we used. Also, since we have P(x) expressed in terms of R(x), maybe we can use interpolation?

But since we already have the answer through two methods, and all checks out, I think 15 is correct. However, just to be thorough, let me compute P(5) using the polynomial expression:

P(x) = x³ - (5/2)x² - (21/2)x + 5

So P(5) = 125 - (5/2)(25) - (21/2)(5) + 5

Compute each term:

(5/2)(25) = 125/2 = 62.5

(21/2)(5) = 105/2 = 52.5

So:

125 - 62.5 - 52.5 + 5 = 125 - 62.5 = 62.5; 62.5 - 52.5 = 10; 10 + 5 = 15. Yep, same result.

Therefore, I think the answer is 15. But let me check once more if there's a possibility of miscalculations.

Wait, but let me think again. The problem says that P(x) is monic. Our polynomial is x³ - (5/2)x² - (21/2)x + 5. The leading coefficient is 1, so that's correct. So that's okay.

Alternatively, maybe there's a smarter way using the Chinese Remainder Theorem or something? Let me think.

Since P(x) ≡ R(x) mod (x - 1)(x - 4), and P(x) ≡ 2R(x) mod (x - 2)(x - 3). Also, since (x - 1)(x - 4) and (x - 2)(x - 3) are coprime? Wait, are they? Let me check if they have common factors. The roots of (x - 1)(x - 4) are 1 and 4, and roots of (x - 2)(x - 3) are 2 and 3. So they are distinct, so the two quadratics are coprime. Therefore, by Chinese Remainder Theorem, there exists a unique polynomial modulo (x - 1)(x - 4)(x - 2)(x - 3) that satisfies both congruences. However, since P(x) is a cubic, and (x - 1)(x - 2)(x - 3)(x - 4) is quartic, so P(x) is determined uniquely by its values at four points? Wait, but maybe not necessary here.

Alternatively, since we have P(x) - R(x) is divisible by (x - 1)(x - 4), and P(x) - 2R(x) is divisible by (x - 2)(x - 3). Therefore, P(x) - R(x) = (x - 1)(x - 4)A(x), and P(x) - 2R(x) = (x - 2)(x - 3)B(x), where A(x) and B(x) are linear polynomials. But since P(x) is cubic, A(x) and B(x) are linear. But since we already solved it through another method, maybe not necessary.

Alternatively, since we know R(x) = -2x -5, then P(x) = (x - 1)(x - 4)Q(x) - 2x -5. Since P(x) is monic cubic, Q(x) is linear, as we saw. But since we already found Q(x) = x + 5/2, then P(x) = (x - 1)(x - 4)(x + 5/2) - 2x -5. Let me compute this expression for x = 5:

First compute (5 - 1)(5 - 4)(5 + 5/2) - 2*5 -5 = (4)(1)(15/2) - 10 -5 = 4*(15/2) -15 = 30 -15 = 15. Same result. So that's another way to compute P(5) directly. Therefore, 15 is correct.

Therefore, I think the answer is 15. So I can write that as \boxed{15}.

**Final Answer**
\boxed{15}

To solve the problem, we are given that $ P(x) $ is a **monic cubic polynomial**, meaning its leading coefficient is 1. We are also told:

- When $ P(x) $ is divided by $ (x - 1)(x - 4) $, the remainder is $ R(x) $.
- When $ P(x) $ is divided by $ (x - 2)(x - 3) $, the remainder is $ 2R(x) $.
- $ P(0) = 5 $.

We are to find $ P(5) $.

---

### Step 1: Understand the Structure of the Polynomial

Since $ P(x) $ is a cubic polynomial, we can write it in the form:

$$
P(x) = x^3 + ax^2 + bx + c
$$

We are told that $ P(0) = 5 $, so:

$$
P(0) = c = 5
$$

So, the polynomial becomes:

$$
P(x) = x^3 + ax^2 + bx + 5
$$

---

### Step 2: Use the Remainder Theorem

Let’s denote the remainder $ R(x) $ as a linear polynomial:

$$
R(x) = mx + n
$$

Then, from the problem:

- $ P(1) = R(1) = m + n $
- $ P(4) = R(4) = 4m + n $
- $ P(2) = 2R(2) = 2(2m + n) = 4m + 2n $
- $ P(3) = 2R(3) = 2(3m + n) = 6m + 2n $

Now, compute $ P(1), P(2), P(3), P(4) $ using the general form of $ P(x) $:

$$
\begin{aligned}
P(1) &= 1 + a + b + 5 = a + b + 6 \\
P(2) &= 8 + 4a + 2b + 5 = 4a + 2b + 13 \\
P(3) &= 27 + 9a + 3b + 5 = 9a + 3b + 32 \\
P(4) &= 64 + 16a + 4b + 5 = 16a + 4b + 69
\end{aligned}
$$

Now equate these expressions to the ones in terms of $ m $ and $ n $:

$$
\begin{aligned}
a + b + 6 &= m + n \quad \text{(1)} \\
4a + 2b + 13 &= 4m + 2n \quad \text{(2)} \\
9a + 3b + 32 &= 6m + 2n \quad \text{(3)} \\
16a + 4b + 69 &= 4m + n \quad \text{(4)}
\end{aligned}
$$

---

### Step 3: Solve the System of Equations

From equation (1):
$$
m + n = a + b + 6 \quad \text{(1a)}
$$

From equation (2):
$$
4a + 2b + 13 = 4m + 2n \quad \text{(2a)}
$$

From equation (3):
$$
9a + 3b + 32 = 6m + 2n \quad \text{(3a)}
$$

From equation (4):
$$
16a + 4b + 69 = 4m + n \quad \text{(4a)}
$$

Subtract (2a) from (3a):

$$
(9a + 3b + 32) - (4a + 2b + 13) = (6m + 2n) - (4m + 2n) \\
5a + b + 19 = 2m \quad \text{(5)}
$$

Subtract (2a) from (4a):

$$
(16a + 4b + 69) - (4a + 2b + 13) = (4m + n) - (4m + 2n) \\
12a + 2b + 56 = -n \quad \text{(6)}
$$

Now solve equations (5) and (6) together.

From (5):
$$
5a + b = 2m - 19 \quad \text{(5a)}
$$

From (6):
$$
12a + 2b = -n - 56 \quad \text{(6a)}
$$

Substitute (5a) into (6a):

$$
12a + 2(2m - 19 - 5a) = -n - 56 \\
12a + 4m - 38 - 10a = -n - 56 \\
2a + 4m - 38 = -n - 56 \\
2a + 4m + 18 = -n \quad \text{(7)}
$$

Now use (1a):
$$
m + n = a + b + 6 \Rightarrow n = a + b + 6 - m
$$

Substitute into (7):

$$
2a + 4m + 18 = -(a + b + 6 - m) \\
2a + 4m + 18 = -a - b - 6 + m \\
3a + 3m + b + 24 = 0 \quad \text{(8)}
$$

Now use (5a):
$$
5a + b = 2m - 19 \Rightarrow b = 2m - 19 - 5a
$$

Substitute into (8):

$$
3a + 3m + (2m - 19 - 5a) + 24 = 0 \\
3a + 3m + 2m - 19 - 5a + 24 = 0 \\
-2a + 5m + 5 = 0 \Rightarrow -2a + 5m = -5 \quad \text{(9)}
$$

Now solve (9) and (5a):

From (5a):
$$
5a + b = 2m - 19
$$

From (9):
$$
-2a + 5m = -5 \Rightarrow 5m = 2a - 5 \Rightarrow m = \frac{2a - 5}{5}
$$

Substitute into (5a):

$$
5a + b = 2\left(\frac{2a - 5}{5}\right) - 19 = \frac{4a - 10}{5} - 19 = \frac{4a - 10 - 95}{5} = \frac{4a - 105}{5}
$$

Multiply both sides by 5:

$$
25a + 5b = 4a - 105 \Rightarrow 21a + 5b = -105 \quad \text{(10)}
$$

Now solve (10) and (9) together.

From (9):
$$
5m = 2a - 5 \Rightarrow m = \frac{2a - 5}{5}
$$

From (10):
$$
21a + 5b = -105 \Rightarrow b = \frac{-105 - 21a}{5}
$$

Now use (1a):
$$
m + n = a + b + 6
$$

Substitute $ m $ and $ b $:

$$
\frac{2a - 5}{5} + n = a + \frac{-105 - 21a}{5} + 6
$$

Multiply through by 5:

$$
2a - 5 + 5n = 5a - 105 - 21a + 30 \\
2a - 5 + 5n = -16a - 75 \\
5n = -18a - 70 \Rightarrow n = -\frac{18a + 70}{5}
$$

Now plug into (1a):

$$
\frac{2a - 5}{5} + \left(-\frac{18a + 70}{5}\right) = a + \frac{-105 - 21a}{5} + 6
$$

Multiply through by 5:

$$
2a - 5 - 18a - 70 = 5a - 105 - 21a + 30 \\
-16a - 75 = -16a - 75
$$

So the system is consistent. Now solve for $ a $ and $ b $.

From (9):
$$
-2a + 5m = -5 \Rightarrow m = \frac{2a - 5}{5}
$$

From (5a):
$$
5a + b = 2m - 19 = 2\left(\frac{2a - 5}{5}\right) - 19 = \frac{4a - 10}{5} - 19 = \frac{4a - 10 - 95}{5} = \frac{4a - 105}{5}
$$

Multiply both sides by 5:

$$
25a + 5b = 4a - 105 \Rightarrow 21a + 5b = -105
$$

Solve for $ b $:

$$
b = \frac{-105 - 21a}{5}
$$

Now plug into (1a):

$$
m + n = a + b + 6
$$

Substitute $ m = \frac{2a - 5}{5} $, $ b = \frac{-105 - 21a}{5} $, and solve for $ a $.

Eventually, solving the system gives:

$$
a = -\frac{5}{2}, \quad b = -\frac{21}{2}, \quad c = 5
$$

---

### Step 4: Final Polynomial and Evaluation

So the polynomial is:

$$
P(x) = x^3 - \frac{5}{2}x^2 - \frac{21}{2}x + 5
$$

Now compute $ P(5) $:

$$
P(5) = 5^3 - \frac{5}{2}(5^2) - \frac{21}{2}(5) + 5 = 125 - \frac{125}{2} - \frac{105}{2} + 5
$$

$$
= 125 - \frac{230}{2} + 5 = 125 - 115 + 5 = 15
$$

---

### Final Answer

$$
\boxed{15}
$$

Solution:
Let $P(x) = x^3 + ax^2 + bx + 5.$  The remainder $R(x)$ has degree at most 1, so let $R(x) = cx + d.$

When $P(x)$ is divided by $(x - 1)(x - 4),$ the quotient is of the form $x + p,$ so write
\[P(x) = (x + p)(x - 1)(x - 4) + R(x) = (x + p)(x - 1)(x - 4) + cx + d.\]Comparing the coefficients of $x^2,$ we get $a = p - 5.$

When $P(x)$ is divided by $(x - 2)(x - 3),$ the quotient is of the form $x + q,$ so write
\[P(x) = (x + q)(x - 2)(x - 3) + 2R(x) = (x + q)(x - 2)(x - 3) + 2(cx + d).\]Comparing the coefficients of $x^2,$ we get $a = q - 5.$  Hence, $p = q.$

Comparing the coefficients of $x$ in both equations, we get
\begin{align*}
b &= c - 5p + 4, \\
b &= 2c - 5p + 6.
\end{align*}Subtracting these equations, we get $c + 2 = 0,$ so $c = -2.$

Comparing the constant coefficients in the first equation, we get $5 = 4p + d.$  Therefore,
\[P(5) = (5 + p)(4)(1) - 10 + d = 10 + 4p + d = \boxed{15}.\]


Explanation:
Okay, so I need to figure out how many distinct values the function $ f(x) = \left\lfloor \left(-\frac{5}{8}\right)^x \right\rfloor $ can take when $ x $ is in the interval $[0, \infty)$. Alright, let me start by recalling what the floor function does. The floor function $ \lfloor y \rfloor $ gives the greatest integer less than or equal to $ y $. So, for any real number $ y $, $ \lfloor y \rfloor $ is the integer part of $ y $, essentially rounding down to the nearest integer.

Now, the function inside the floor is $ \left(-\frac{5}{8}\right)^x $. Let me think about how this behaves as $ x $ varies from 0 to infinity. First, let me note that $ -\frac{5}{8} $ is a negative number with absolute value less than 1. So, raising it to different powers will have different behaviors depending on whether the exponent is an integer or not. But since $ x $ is in $[0, \infty)$, it can be any non-negative real number, not just integers. Hmm, so I need to consider real exponents here.

Wait, but when dealing with exponents, if the base is negative, we have to be careful. For example, $ (-a)^x $ is only real if $ x $ is an integer or if $ a $ is positive and $ x $ is rational with an odd denominator, etc. But since $ x $ is a real number here, and we are dealing with $ (-\frac{5}{8})^x $, this expression is only real if $ x $ is an integer? Wait, no. Wait, actually, for real exponents, if the base is negative, the expression is not real unless the exponent is rational with an odd denominator. But since $ x $ is any real number in $[0, \infty)$, this might not be defined for all real numbers. Wait, but the problem says that the function is defined for all $ x $ in $[0, \infty)$ such that $ f(x) $ is a real number. So maybe the problem is assuming that $ x $ is such that $ (-\frac{5}{8})^x $ is real? Or maybe there's a misunderstanding here.

Wait, hold on. Let me check. If $ x $ is a real number, then $ (-\frac{5}{8})^x $ is not real unless $ x $ is an integer. Because if $ x $ is not an integer, then you can't take a real root of a negative number. For example, $ (-\frac{5}{8})^{1/2} $ is not real. Therefore, maybe the problem is actually considering only integer values of $ x $? But the problem says "defined for all values of $ x $ in $[0, \infty)$ such that $ f(x) $ is a real number." So perhaps the domain is actually only the integers? Wait, but the problem says "defined for all values of $ x $ in $[0, \infty)$ such that $ f(x) $ is a real number." So maybe the function is defined only for those $ x $ where $ (-\frac{5}{8})^x $ is real. Which would be when $ x $ is an integer? Because otherwise, for non-integer exponents, it's not real. Therefore, maybe the problem is actually considering $ x $ as integers? But the problem says "for all values of $ x $ in $[0, \infty)$", so maybe there's a different interpretation.

Wait, perhaps the problem is using complex numbers? But the floor function is defined for real numbers. So if the expression inside the floor is complex, then the floor function isn't defined. Therefore, the problem must be considering only those $ x $ for which $ (-\frac{5}{8})^x $ is real. Which, as I thought, would be when $ x $ is an integer. Because for non-integer exponents, you can't have a real result. Therefore, maybe the problem is actually considering $ x $ as non-negative integers? But the problem says "for all values of $ x $ in $[0, \infty)$", so maybe there's a mistake here? Or perhaps the problem is using a different definition?

Wait, maybe I need to check if there's a way to define $ (-\frac{5}{8})^x $ for real exponents. Let me recall that for negative bases, raising to real exponents is not generally defined in real numbers. However, if we consider complex numbers, it's possible, but then the floor function isn't defined for complex numbers. Therefore, the problem must be considering only integer exponents. Therefore, maybe the problem is actually about integer values of $ x $, even though it says $ x $ is in $[0, \infty)$. Maybe it's a typo or misunderstanding. Alternatively, maybe the problem is considering $ x $ as real numbers, but the expression is defined in some way?

Alternatively, perhaps the problem is using the principal value or something? Wait, but even so, for non-integer exponents, the expression would be complex. Therefore, unless there's some other interpretation, I think the problem must be considering $ x $ as non-negative integers. Therefore, maybe the problem is actually about integer exponents. Let me check the original problem again.

The problem says: "Let $ f(x) = \left\lfloor \left(-\frac{5}{8}\right)^x \right\rfloor $ be a function that is defined for all values of $ x $ in $[0, \infty)$ such that $ f(x) $ is a real number. How many distinct values exist in the range of $ f(x) $?"

So, the function is defined for all $ x $ in $[0, \infty)$ such that $ f(x) $ is real. Therefore, the domain is the set of $ x \in [0, \infty) $ for which $ (-\frac{5}{8})^x $ is real. Therefore, as discussed, this would be when $ x $ is an integer. Because for non-integer exponents, $ (-\frac{5}{8})^x $ is not real. Therefore, the domain is actually the non-negative integers. Therefore, the problem is effectively considering $ x $ as non-negative integers. Therefore, the function is defined for $ x = 0, 1, 2, 3, \ldots $, and we need to find the number of distinct values of $ \lfloor (-\frac{5}{8})^x \rfloor $ for these integer values of $ x $.

Alternatively, maybe there's a way to define $ (-\frac{5}{8})^x $ for real $ x $, but I don't think so. For example, if $ x $ is rational with an odd denominator, then it's possible, but for irrational exponents, it's not. Therefore, since the problem says "for all values of $ x $ in $[0, \infty)$ such that $ f(x) $ is a real number", the domain is actually the set of $ x $ where $ (-\frac{5}{8})^x $ is real. Which, as I said, is only when $ x $ is an integer. Therefore, the problem is about integer exponents. Therefore, I can proceed under that assumption.

Therefore, let me consider $ x $ as non-negative integers. So $ x = 0, 1, 2, 3, \ldots $. Let me compute $ (-\frac{5}{8})^x $ for these values and then take the floor of them.

Starting with $ x = 0 $: $ (-\frac{5}{8})^0 = 1 $. Therefore, $ f(0) = \lfloor 1 \rfloor = 1 $.

For $ x = 1 $: $ (-\frac{5}{8})^1 = -\frac{5}{8} $. Therefore, $ f(1) = \lfloor -\frac{5}{8} \rfloor $. The floor of a negative number is the next lower integer. Since $ -\frac{5}{8} $ is between -1 and 0, the floor is -1. Therefore, $ f(1) = -1 $.

For $ x = 2 $: $ (-\frac{5}{8})^2 = \left(\frac{25}{64}\right) $. Therefore, $ f(2) = \lfloor \frac{25}{64} \rfloor = 0 $, since $ \frac{25}{64} $ is between 0 and 1.

For $ x = 3 $: $ (-\frac{5}{8})^3 = -\frac{125}{512} $. Therefore, $ f(3) = \lfloor -\frac{125}{512} \rfloor $. Again, this is between -1 and 0, so floor is -1.

For $ x = 4 $: $ (-\frac{5}{8})^4 = \left(\frac{625}{4096}\right) $. So floor is 0.

Similarly, for $ x = 5 $: $ (-\frac{5}{8})^5 = -\frac{3125}{32768} $, floor is -1.

So, it seems like for even exponents, the value is positive and between 0 and 1, so floor is 0. For odd exponents, the value is negative and between -1 and 0, so floor is -1. Except for x=0, which is 1. Therefore, the function f(x) takes values 1, -1, and 0. Wait, but let me check for higher exponents.

Wait, for x=0: 1, floor is 1.

x=1: -5/8 ≈ -0.625, floor is -1.

x=2: 25/64 ≈ 0.390625, floor is 0.

x=3: -125/512 ≈ -0.244, floor is -1.

x=4: 625/4096 ≈ 0.15258789, floor is 0.

x=5: -3125/32768 ≈ -0.095, floor is -1.

x=6: (5/8)^6 ≈ 0.0586, floor is 0.

x=7: - (5/8)^7 ≈ -0.036, floor is -1.

So, as x increases, the magnitude of the value decreases, approaching zero. So for even x, the value is positive, approaching zero from above, so floor is 0. For odd x, the value is negative, approaching zero from below, so floor is -1. Except for x=0, which is 1. Therefore, the function f(x) takes three values: 1, 0, -1. Therefore, the range has three distinct values. Therefore, the answer is 3?

Wait, but let me check if there are any other values. For example, is there any x where the value inside the floor is an integer? For example, when is $ (-\frac{5}{8})^x $ an integer? Well, since $ |\frac{5}{8}| < 1 $, raising it to any positive power will give a number between 0 and 1 (for even exponents) or between -1 and 0 (for odd exponents). Therefore, the only time when $ (-\frac{5}{8})^x $ is an integer is when x=0, which gives 1. For all other x, the value is between -1 and 1, but not equal to any integer except 0 and 1. Wait, but for even exponents, the value is between 0 and 1, so floor is 0. For odd exponents, between -1 and 0, floor is -1. So the only integer values achieved are 1, 0, -1. Therefore, the range is { -1, 0, 1 }, so three distinct values. Therefore, the answer is 3?

But wait, let me check for x=0. Is x=0 included? The problem says x is in [0, ∞). So x=0 is included. So f(0)=1, f(1)=-1, f(2)=0, f(3)=-1, etc. So yes, only three values. Therefore, the answer is 3? But let me check again.

Wait, but maybe there's a case where the value inside the floor is exactly an integer? For example, if $ (-\frac{5}{8})^x = 0 $, but that would require x approaching infinity, but it never actually reaches zero. Similarly, for negative numbers, it approaches zero from below. So the value never actually reaches -1 or 1 except at x=0. Therefore, the floor function will only take -1, 0, 1. Therefore, three distinct values. Therefore, the answer is 3?

But wait, let me check for x=0. The value is 1, floor is 1. For x=1, it's -5/8, floor is -1. For x=2, 25/64, floor is 0. For x=3, -125/512, floor is -1. For x=4, 625/4096, floor is 0. So yes, only three values. Therefore, the answer is 3? But wait, the problem says "distinct values exist in the range of f(x)". So maybe 3?

But wait, let me think again. Wait, is there a possibility that for some x, the value inside the floor is exactly an integer? For example, if $ (-\frac{5}{8})^x = 0 $, but that's not possible. If $ (-\frac{5}{8})^x = 1 $, then x=0. If $ (-\frac{5}{8})^x = -1 $, then we need $ (-\frac{5}{8})^x = -1 $. Let me check if that's possible. Let me solve for x:

$ (-\frac{5}{8})^x = -1 $.

Taking natural logarithm on both sides:

$ x \ln(-\frac{5}{8}) = \ln(-1) $.

But ln(-1) is not defined in real numbers. Therefore, there is no real solution for x. Therefore, the equation $ (-\frac{5}{8})^x = -1 $ has no real solution. Therefore, the value inside the floor is never exactly -1 or 1 except at x=0. Therefore, the floor function will only take -1, 0, 1. Therefore, three distinct values. Therefore, the answer is 3?

But wait, let me check for x=0. The value is 1, floor is 1. For x=1, floor is -1. For x=2, floor is 0. For x=3, floor is -1. For x=4, floor is 0. So yes, only three values. Therefore, the answer is 3? But I need to make sure that there are no other values. For example, is there any x where the value inside the floor is, say, -2? Let me check. For example, if $ (-\frac{5}{8})^x $ is less than -1, then floor would be less than or equal to -2. But since $ (-\frac{5}{8})^x $ for x positive is between -1 and 1 (excluding -1 and 1 except at x=0). Wait, for x positive, the absolute value is $ (\frac{5}{8})^x $, which is less than 1. Therefore, $ (-\frac{5}{8})^x $ is between -1 and 1. Therefore, the floor function can only be -1, 0, or 1. Therefore, only three values. Therefore, the answer is 3. Therefore, the number of distinct values is 3.

But wait, let me check for x=0. Is x=0 included? Yes. So f(0)=1. Then for x=1, f(x)=-1. For x=2, 0. For x=3, -1. For x=4, 0. So the range is { -1, 0, 1 }, so three distinct values. Therefore, the answer is 3. Therefore, the answer is \boxed{3}?

Wait, but let me check again. Maybe there's a mistake here. Let me think again. Suppose x is a non-integer. Wait, but earlier I thought that the function is only defined for integer x. But the problem says "defined for all values of x in [0, ∞) such that f(x) is a real number". So if x is not an integer, then $ (-\frac{5}{8})^x $ is not real. Therefore, the domain is only integers. Therefore, the function is defined only for integer x. Therefore, the range is { -1, 0, 1 }, so three values. Therefore, answer is 3. But maybe I need to check if there are other integer values?

Wait, for x=0, 1, 2, 3,... So for x=0, 1, 2,..., the function takes values 1, -1, 0, -1, 0,... So only three distinct values. Therefore, answer is 3. Therefore, I think the answer is 3. But let me check again.

Alternatively, maybe the problem is considering x as real numbers, and using some complex analysis? But then floor function is not defined. Therefore, the problem must be considering integer exponents. Therefore, answer is 3. Therefore, I think the answer is 3. However, let me check once more.

Wait, but maybe there's a case where for some non-integer x, the expression is real? For example, if x is a rational number with odd denominator? Let me think. Suppose x is a rational number, say x = p/q, where p and q are integers, q > 0, and p and q coprime. Then, $ (-\frac{5}{8})^{p/q} $ is real if and only if the denominator q is odd. Because if q is odd, then you can take the q-th root of a negative number. For example, $ (-\frac{5}{8})^{1/3} $ is real. However, if q is even, then it's not real. Therefore, if x is a rational number with odd denominator, then the expression is real. However, if x is irrational, then it's not real. Therefore, the domain of the function is all real numbers x in [0, ∞) such that x can be expressed as a rational number with odd denominator. However, this is a very complicated set, and the problem says "defined for all values of x in [0, ∞) such that f(x) is a real number". Therefore, the domain is all x in [0, ∞) for which $ (-\frac{5}{8})^x $ is real. Which includes integers (since they can be written as p/1, denominator 1, which is odd), and other rationals with odd denominators, and irrationals? No, irrationals can't be expressed as p/q with integer p, q. Therefore, the domain is all real numbers x in [0, ∞) that are rational numbers with odd denominators when expressed in lowest terms. However, this is a complicated domain. Therefore, the problem is not straightforward. Therefore, maybe the problem is intended to consider only integer exponents? Because otherwise, the function is defined on a complicated set, and the range would be more complex.

But the problem says "defined for all values of x in [0, ∞) such that f(x) is a real number". Therefore, the domain is all x in [0, ∞) where $ (-\frac{5}{8})^x $ is real. Therefore, as discussed, this includes integers and some other rationals. However, if we consider that, then the function is defined on a dense set in [0, ∞), but the problem is asking for the number of distinct values in the range. Therefore, maybe we need to analyze the function over all real numbers x where it is defined.

But this seems complicated. However, maybe the problem is intended to consider only integer exponents. Because otherwise, the problem becomes more complex. Let me check the original problem again. It says "defined for all values of x in [0, ∞) such that f(x) is a real number". Therefore, the function is defined on the set of x in [0, ∞) where $ (-\frac{5}{8})^x $ is real. Therefore, as discussed, this is the set of x which are rational numbers with odd denominators. However, even so, the function's range would depend on the values of x in that domain.

But this seems too complicated for a problem that is likely expecting an answer of 3. Therefore, maybe the problem is intended to consider x as integers. Therefore, given that, the answer is 3. However, to be thorough, let me consider the case where x is allowed to be any real number where $ (-\frac{5}{8})^x $ is real. Let me think about how the function behaves.

First, for x rational with odd denominator, say x = p/q, where q is odd. Then, $ (-\frac{5}{8})^{p/q} = \left( (-\frac{5}{8})^{1/q} \right)^p $. Since q is odd, $ (-\frac{5}{8})^{1/q} $ is real. Let me denote $ a = (-\frac{5}{8})^{1/q} $. Then, $ a $ is a real number. Since $ -\frac{5}{8} $ is negative, and q is odd, $ a $ is negative. Therefore, $ a = -|a| $, where $ |a| = (\frac{5}{8})^{1/q} $. Therefore, $ (-\frac{5}{8})^{p/q} = (-1)^p \cdot (\frac{5}{8})^{p/q} $. Therefore, depending on whether p is even or odd, the value is positive or negative. Therefore, for x rational with odd denominator, the value $ (-\frac{5}{8})^x $ is either positive or negative, depending on the parity of the numerator p. However, since x can be any such rational number, the value $ (-\frac{5}{8})^x $ can be positive or negative, and its magnitude is $ (\frac{5}{8})^x $, which is between 0 and 1 for x > 0. Therefore, similar to the integer case, for x > 0, the value is between -1 and 1, excluding -1 and 1. Therefore, the floor function would be -1, 0, or 1. However, for x=0, it's 1. Therefore, even if we consider x as real numbers where the expression is defined, the floor function would still only take values -1, 0, 1. Therefore, the range is still { -1, 0, 1 }, so three distinct values. Therefore, regardless of whether x is integer or rational with odd denominator, the range is still three values. Therefore, the answer is 3.

But wait, let me check for some specific non-integer x. For example, take x = 1/3. Then $ (-\frac{5}{8})^{1/3} = - (\frac{5}{8})^{1/3} \approx -0.855 $. Therefore, floor of that is -1. Similarly, x = 2/3: $ (-\frac{5}{8})^{2/3} = [(-\frac{5}{8})^{1/3}]^2 = [ - (\frac{5}{8})^{1/3} ]^2 = (\frac{5}{8})^{2/3} \approx 0.63 $. Therefore, floor is 0. Similarly, x = 1/1 (which is integer) gives -5/8 ≈ -0.625, floor -1. x = 2/1 = 2, gives 25/64 ≈ 0.39, floor 0. So even for non-integer x, the floor function still gives -1, 0, or 1. Therefore, regardless of x being integer or non-integer (as long as the expression is real), the floor function only takes those three values. Therefore, the range is { -1, 0, 1 }, so three distinct values. Therefore, the answer is 3.

But wait, let me check for x approaching infinity. As x approaches infinity, $ (-\frac{5}{8})^x $ approaches 0, since $ |\frac{5}{8}| < 1 $. Therefore, for x approaching infinity, the value approaches 0. Therefore, for x large enough, the value is between -1 and 1, but very close to 0. Therefore, floor is -1 or 0 depending on the sign. Therefore, even for non-integer x, the floor function is still -1 or 0. Therefore, the range is still { -1, 0, 1 }.

Therefore, regardless of the domain (as long as it's the set of x where the expression is real), the range has three elements. Therefore, the answer is 3. Therefore, the number of distinct values in the range of f(x) is 3. Therefore, the answer is \boxed{3}.

But wait, let me check once more. Suppose x is a non-integer rational with odd denominator. For example, x = 1/3, which we did. The value is approximately -0.855, floor is -1. If x is 1/5, then $ (-\frac{5}{8})^{1/5} \approx - (0.625)^{1/5} \approx -0.92 $, floor is -1. If x is 1/1000, then $ (-\frac{5}{8})^{1/1000} $ is a negative number very close to -1, since raising to a small exponent. Wait, no. Wait, $ (-\frac{5}{8})^{1/n} $ as n approaches infinity. Let me think. For x approaching 0 from the positive side, $ (-\frac{5}{8})^x $. Wait, if x approaches 0, then $ (-\frac{5}{8})^x $ approaches 1. For example, x approaching 0, even if x is a non-integer. For example, x approaching 0 from the positive side, say x = 0.1. Then $ (-\frac{5}{8})^{0.1} $. Wait, but this is not real. Wait, if x is 0.1, which is not rational with odd denominator? Wait, 0.1 is 1/10, which has denominator 10, which is even. Therefore, $ (-\frac{5}{8})^{0.1} $ is not real. Therefore, x=0.1 is not in the domain. Therefore, the domain is only those x which are rational numbers with odd denominators. Therefore, for example, x=1/3, 1/5, 1/7, etc. So, for x approaching 0 from the right, with x being rational with odd denominator, say x = 1/n where n is odd. Then, $ (-\frac{5}{8})^{1/n} $ is a negative number approaching -1 as n approaches infinity? Wait, no. Wait, as n approaches infinity, $ (-\frac{5}{8})^{1/n} $ approaches 1? Wait, no. Let me think. Let me take the limit as n approaches infinity of $ (-\frac{5}{8})^{1/n} $. Let me write it as $ e^{\frac{1}{n} \ln(-\frac{5}{8})} $. But ln(-5/8) is not real. Therefore, this is not defined. Wait, but earlier we said that for x rational with odd denominator, the expression is real. So for x = 1/n, with n odd, then $ (-\frac{5}{8})^{1/n} $ is real. Let me compute the limit as n approaches infinity of $ (-\frac{5}{8})^{1/n} $. Let me write it as $ (-1)^{1/n} \cdot (\frac{5}{8})^{1/n} $. Since n is odd, $ (-1)^{1/n} = -1 $. Therefore, the expression is $ - (\frac{5}{8})^{1/n} $. As n approaches infinity, $ (\frac{5}{8})^{1/n} $ approaches 1, so the whole expression approaches -1. Therefore, as x approaches 0 from the right (with x being rational with odd denominator), the value approaches -1. Therefore, for x approaching 0 from the right, the value approaches -1. Therefore, for x very close to 0, the value is slightly greater than -1? Wait, no. Wait, if x approaches 0 from the right, then $ (-\frac{5}{8})^x $ approaches 1? Wait, no. Wait, if x approaches 0, then for x positive, $ (-\frac{5}{8})^x $ is... Wait, if x is approaching 0, but x is rational with odd denominator. For example, take x = 1/n, n odd, approaching 0. Then, as n increases, x approaches 0. Then, $ (-\frac{5}{8})^{1/n} = - (\frac{5}{8})^{1/n} $. As n approaches infinity, $ (\frac{5}{8})^{1/n} $ approaches 1, so the whole expression approaches -1. Therefore, for x approaching 0 from the right, the value approaches -1. Therefore, for x very small positive, the value is slightly greater than -1? Wait, no. Wait, $ (\frac{5}{8})^{1/n} $ is less than 1 for n > 0. Therefore, $ - (\frac{5}{8})^{1/n} $ is greater than -1. Therefore, as x approaches 0 from the right, the value approaches -1 from above. Therefore, for x approaching 0, the value is between -1 and 0, approaching -1. Therefore, floor of that is -1. Therefore, even for x approaching 0, floor is -1. Except when x=0, which is 1. Therefore, the floor function is -1 for all x > 0, except when the value is between 0 and 1, which occurs when x is even? Wait, no. Wait, for x being rational with odd denominator, the value is negative or positive depending on whether the exponent is even or odd? Wait, earlier we saw that $ (-\frac{5}{8})^x = (-1)^p \cdot (\frac{5}{8})^x $, where x = p/q with q odd. Wait, no, earlier we had x = p/q, then $ (-\frac{5}{8})^{p/q} = (-1)^p \cdot (\frac{5}{8})^{p/q} $. Therefore, if p is even, the value is positive; if p is odd, the value is negative. Therefore, for x rational with odd denominator, depending on the numerator's parity, the value is positive or negative. Therefore, for example, if x is 2/3 (p=2, q=3), then the value is positive. If x is 1/3, it's negative. Therefore, for x rational with odd denominator, the value can be positive or negative. Therefore, for positive x, depending on whether the exponent is even or odd (in terms of numerator), the value is positive or negative. Therefore, for x positive, the value is between -1 and 1, and floor is -1, 0, or 1. However, when is the value between 0 and 1? When the exponent is even (in terms of numerator). For example, x = 2/3, which is even numerator, gives positive value between 0 and 1. Similarly, x = 4/5, which is even numerator, gives positive value between 0 and 1. Therefore, floor is 0. For x with odd numerator, the value is negative between -1 and 0, floor is -1. For x=0, floor is 1. Therefore, the range is still { -1, 0, 1 }, so three distinct values. Therefore, regardless of x being integer or non-integer (as long as the expression is real), the range is three values. Therefore, the answer is 3.

Therefore, I think the answer is 3. Therefore, the number of distinct values in the range of f(x) is \boxed{3}.

**Final Answer**
\boxed{3}

To determine how many **distinct values** exist in the range of the function
$$
f(x) = \left\lfloor \left(-\frac{5}{8}\right)^x \right\rfloor
$$
for $ x \in [0, \infty) $, we must first understand the behavior of the expression inside the floor function:
$$
\left(-\frac{5}{8}\right)^x
$$

---

### Step 1: Understanding the Domain

The base $ -\frac{5}{8} $ is a **negative real number** with absolute value less than 1. When raising a negative number to a real exponent, the result is **not always real**. Specifically:

- If $ x $ is an **integer**, then $ \left(-\frac{5}{8}\right)^x $ is real.
- If $ x $ is a **non-integer rational number** with an **odd denominator**, then the expression is also real (since we can take an odd root of a negative number).
- If $ x $ is an **irrational number**, the expression is **not real**.

However, the problem states that the function is defined for all $ x \in [0, \infty) $ such that $ f(x) $ is a real number. Therefore, the domain of $ f(x) $ is the set of all $ x \in [0, \infty) $ for which $ \left(-\frac{5}{8}\right)^x $ is real. This includes:

- All **non-negative integers** $ x = 0, 1, 2, \dots $
- All **non-integer rational numbers** with **odd denominators** in their reduced form.

But for the purpose of determining the **range**, we can consider the behavior of the function over the entire domain where it is defined.

---

### Step 2: Behavior of the Function

Let’s analyze the expression $ \left(-\frac{5}{8}\right)^x $ for different values of $ x $:

- **When $ x = 0 $:**
  $$
  \left(-\frac{5}{8}\right)^0 = 1 \Rightarrow f(0) = \lfloor 1 \rfloor = 1
  $$

- **When $ x $ is a **positive integer** (odd or even):**
  - If $ x $ is **odd**, then $ \left(-\frac{5}{8}\right)^x $ is **negative** and lies in the interval $ (-1, 0) $, so
    $$
    f(x) = \lfloor \text{negative number in } (-1, 0) \rfloor = -1
    $$
  - If $ x $ is **even**, then $ \left(-\frac{5}{8}\right)^x $ is **positive** and lies in the interval $ (0, 1) $, so
    $$
    f(x) = \lfloor \text{positive number in } (0, 1) \rfloor = 0
    $$

- **When $ x $ is a **non-integer rational number** with an **odd denominator** (e.g., $ x = \frac{1}{3}, \frac{2}{5}, \dots $):**
  - If the numerator is **odd**, the result is **negative** and in $ (-1, 0) $, so $ f(x) = -1 $
  - If the numerator is **even**, the result is **positive** and in $ (0, 1) $, so $ f(x) = 0 $

- **As $ x \to \infty $:**
  The magnitude of $ \left(-\frac{5}{8}\right)^x $ approaches 0, so the value is always in $ (-1, 1) $, and the floor function still yields either $ -1 $, $ 0 $, or $ 1 $.

---

### Step 3: Determining the Range

From the above, we observe that the function $ f(x) $ can only take the following **three distinct values**:

- $ f(x) = 1 $ when $ x = 0 $
- $ f(x) = -1 $ when $ x $ is a **positive rational number with odd numerator** (i.e., odd exponent)
- $ f(x) = 0 $ when $ x $ is a **positive rational number with even numerator** (i.e., even exponent)

Thus, the **range** of $ f(x) $ is:
$$
\{ -1, 0, 1 \}
$$

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many distinct values the function $ f(x) = \left\lfloor \left(-\frac{5}{8}\right)^x \right\rfloor $ can take when $ x $ is in the interval $[0, \infty)$. Alright, let me start by recalling what the floor function does. The floor function $ \lfloor y \rfloor $ gives the greatest integer less than or equal to $ y $. So, for any real number $ y $, $ \lfloor y \rfloor $ is the integer part of $ y $, essentially rounding down to the nearest integer.

Now, the function inside the floor is $ \left(-\frac{5}{8}\right)^x $. Let me think about how this behaves as $ x $ varies from 0 to infinity. First, let me note that $ -\frac{5}{8} $ is a negative number with absolute value less than 1. So, raising it to different powers will have different behaviors depending on whether the exponent is an integer or not. But since $ x $ is in $[0, \infty)$, it can be any non-negative real number, not just integers. Hmm, so I need to consider real exponents here.

Wait, but when dealing with exponents, if the base is negative, we have to be careful. For example, $ (-a)^x $ is only real if $ x $ is an integer or if $ a $ is positive and $ x $ is rational with an odd denominator, etc. But since $ x $ is a real number here, and we are dealing with $ (-\frac{5}{8})^x $, this expression is only real if $ x $ is an integer? Wait, no. Wait, actually, for real exponents, if the base is negative, the expression is not real unless the exponent is rational with an odd denominator. But since $ x $ is any real number in $[0, \infty)$, this might not be defined for all real numbers. Wait, but the problem says that the function is defined for all $ x $ in $[0, \infty)$ such that $ f(x) $ is a real number. So maybe the problem is assuming that $ x $ is such that $ (-\frac{5}{8})^x $ is real? Or maybe there's a misunderstanding here.

Wait, hold on. Let me check. If $ x $ is a real number, then $ (-\frac{5}{8})^x $ is not real unless $ x $ is an integer. Because if $ x $ is not an integer, then you can't take a real root of a negative number. For example, $ (-\frac{5}{8})^{1/2} $ is not real. Therefore, maybe the problem is actually considering only integer values of $ x $? But the problem says "defined for all values of $ x $ in $[0, \infty)$ such that $ f(x) $ is a real number." So perhaps the domain is actually only the integers? Wait, but the problem says "defined for all values of $ x $ in $[0, \infty)$ such that $ f(x) $ is a real number." So maybe the function is defined only for those $ x $ where $ (-\frac{5}{8})^x $ is real. Which would be when $ x $ is an integer? Because otherwise, for non-integer exponents, it's not real. Therefore, maybe the problem is actually considering $ x $ as integers? But the problem says "for all values of $ x $ in $[0, \infty)$", so maybe there's a different interpretation.

Wait, perhaps the problem is using complex numbers? But the floor function is defined for real numbers. So if the expression inside the floor is complex, then the floor function isn't defined. Therefore, the problem must be considering only those $ x $ for which $ (-\frac{5}{8})^x $ is real. Which, as I thought, would be when $ x $ is an integer. Because for non-integer exponents, you can't have a real result. Therefore, maybe the problem is actually considering $ x $ as non-negative integers? But the problem says "for all values of $ x $ in $[0, \infty)$", so maybe there's a mistake here? Or perhaps the problem is using a different definition?

Wait, maybe I need to check if there's a way to define $ (-\frac{5}{8})^x $ for real exponents. Let me recall that for negative bases, raising to real exponents is not generally defined in real numbers. However, if we consider complex numbers, it's possible, but then the floor function isn't defined for complex numbers. Therefore, the problem must be considering only integer exponents. Therefore, maybe the problem is actually about integer values of $ x $, even though it says $ x $ is in $[0, \infty)$. Maybe it's a typo or misunderstanding. Alternatively, maybe the problem is considering $ x $ as real numbers, but the expression is defined in some way?

Alternatively, perhaps the problem is using the principal value or something? Wait, but even so, for non-integer exponents, the expression would be complex. Therefore, unless there's some other interpretation, I think the problem must be considering $ x $ as non-negative integers. Therefore, maybe the problem is actually about integer exponents. Let me check the original problem again.

The problem says: "Let $ f(x) = \left\lfloor \left(-\frac{5}{8}\right)^x \right\rfloor $ be a function that is defined for all values of $ x $ in $[0, \infty)$ such that $ f(x) $ is a real number. How many distinct values exist in the range of $ f(x) $?"

So, the function is defined for all $ x $ in $[0, \infty)$ such that $ f(x) $ is real. Therefore, the domain is the set of $ x \in [0, \infty) $ for which $ (-\frac{5}{8})^x $ is real. Therefore, as discussed, this would be when $ x $ is an integer. Because for non-integer exponents, $ (-\frac{5}{8})^x $ is not real. Therefore, the domain is actually the non-negative integers. Therefore, the problem is effectively considering $ x $ as non-negative integers. Therefore, the function is defined for $ x = 0, 1, 2, 3, \ldots $, and we need to find the number of distinct values of $ \lfloor (-\frac{5}{8})^x \rfloor $ for these integer values of $ x $.

Alternatively, maybe there's a way to define $ (-\frac{5}{8})^x $ for real $ x $, but I don't think so. For example, if $ x $ is rational with an odd denominator, then it's possible, but for irrational exponents, it's not. Therefore, since the problem says "for all values of $ x $ in $[0, \infty)$ such that $ f(x) $ is a real number", the domain is actually the set of $ x $ where $ (-\frac{5}{8})^x $ is real. Which, as I said, is only when $ x $ is an integer. Therefore, the problem is about integer exponents. Therefore, I can proceed under that assumption.

Therefore, let me consider $ x $ as non-negative integers. So $ x = 0, 1, 2, 3, \ldots $. Let me compute $ (-\frac{5}{8})^x $ for these values and then take the floor of them.

Starting with $ x = 0 $: $ (-\frac{5}{8})^0 = 1 $. Therefore, $ f(0) = \lfloor 1 \rfloor = 1 $.

For $ x = 1 $: $ (-\frac{5}{8})^1 = -\frac{5}{8} $. Therefore, $ f(1) = \lfloor -\frac{5}{8} \rfloor $. The floor of a negative number is the next lower integer. Since $ -\frac{5}{8} $ is between -1 and 0, the floor is -1. Therefore, $ f(1) = -1 $.

For $ x = 2 $: $ (-\frac{5}{8})^2 = \left(\frac{25}{64}\right) $. Therefore, $ f(2) = \lfloor \frac{25}{64} \rfloor = 0 $, since $ \frac{25}{64} $ is between 0 and 1.

For $ x = 3 $: $ (-\frac{5}{8})^3 = -\frac{125}{512} $. Therefore, $ f(3) = \lfloor -\frac{125}{512} \rfloor $. Again, this is between -1 and 0, so floor is -1.

For $ x = 4 $: $ (-\frac{5}{8})^4 = \left(\frac{625}{4096}\right) $. So floor is 0.

Similarly, for $ x = 5 $: $ (-\frac{5}{8})^5 = -\frac{3125}{32768} $, floor is -1.

So, it seems like for even exponents, the value is positive and between 0 and 1, so floor is 0. For odd exponents, the value is negative and between -1 and 0, so floor is -1. Except for x=0, which is 1. Therefore, the function f(x) takes values 1, -1, and 0. Wait, but let me check for higher exponents.

Wait, for x=0: 1, floor is 1.

x=1: -5/8 ≈ -0.625, floor is -1.

x=2: 25/64 ≈ 0.390625, floor is 0.

x=3: -125/512 ≈ -0.244, floor is -1.

x=4: 625/4096 ≈ 0.15258789, floor is 0.

x=5: -3125/32768 ≈ -0.095, floor is -1.

x=6: (5/8)^6 ≈ 0.0586, floor is 0.

x=7: - (5/8)^7 ≈ -0.036, floor is -1.

So, as x increases, the magnitude of the value decreases, approaching zero. So for even x, the value is positive, approaching zero from above, so floor is 0. For odd x, the value is negative, approaching zero from below, so floor is -1. Except for x=0, which is 1. Therefore, the function f(x) takes three values: 1, 0, -1. Therefore, the range has three distinct values. Therefore, the answer is 3?

Wait, but let me check if there are any other values. For example, is there any x where the value inside the floor is an integer? For example, when is $ (-\frac{5}{8})^x $ an integer? Well, since $ |\frac{5}{8}| < 1 $, raising it to any positive power will give a number between 0 and 1 (for even exponents) or between -1 and 0 (for odd exponents). Therefore, the only time when $ (-\frac{5}{8})^x $ is an integer is when x=0, which gives 1. For all other x, the value is between -1 and 1, but not equal to any integer except 0 and 1. Wait, but for even exponents, the value is between 0 and 1, so floor is 0. For odd exponents, between -1 and 0, floor is -1. So the only integer values achieved are 1, 0, -1. Therefore, the range is { -1, 0, 1 }, so three distinct values. Therefore, the answer is 3?

But wait, let me check for x=0. Is x=0 included? The problem says x is in [0, ∞). So x=0 is included. So f(0)=1, f(1)=-1, f(2)=0, f(3)=-1, etc. So yes, only three values. Therefore, the answer is 3? But let me check again.

Wait, but maybe there's a case where the value inside the floor is exactly an integer? For example, if $ (-\frac{5}{8})^x = 0 $, but that would require x approaching infinity, but it never actually reaches zero. Similarly, for negative numbers, it approaches zero from below. So the value never actually reaches -1 or 1 except at x=0. Therefore, the floor function will only take -1, 0, 1. Therefore, three distinct values. Therefore, the answer is 3?

But wait, let me check for x=0. The value is 1, floor is 1. For x=1, it's -5/8, floor is -1. For x=2, 25/64, floor is 0. For x=3, -125/512, floor is -1. For x=4, 625/4096, floor is 0. So yes, only three values. Therefore, the answer is 3? But wait, the problem says "distinct values exist in the range of f(x)". So maybe 3?

But wait, let me think again. Wait, is there a possibility that for some x, the value inside the floor is exactly an integer? For example, if $ (-\frac{5}{8})^x = 0 $, but that's not possible. If $ (-\frac{5}{8})^x = 1 $, then x=0. If $ (-\frac{5}{8})^x = -1 $, then we need $ (-\frac{5}{8})^x = -1 $. Let me check if that's possible. Let me solve for x:

$ (-\frac{5}{8})^x = -1 $.

Taking natural logarithm on both sides:

$ x \ln(-\frac{5}{8}) = \ln(-1) $.

But ln(-1) is not defined in real numbers. Therefore, there is no real solution for x. Therefore, the equation $ (-\frac{5}{8})^x = -1 $ has no real solution. Therefore, the value inside the floor is never exactly -1 or 1 except at x=0. Therefore, the floor function will only take -1, 0, 1. Therefore, three distinct values. Therefore, the answer is 3?

But wait, let me check for x=0. The value is 1, floor is 1. For x=1, floor is -1. For x=2, floor is 0. For x=3, floor is -1. For x=4, floor is 0. So yes, only three values. Therefore, the answer is 3? But I need to make sure that there are no other values. For example, is there any x where the value inside the floor is, say, -2? Let me check. For example, if $ (-\frac{5}{8})^x $ is less than -1, then floor would be less than or equal to -2. But since $ (-\frac{5}{8})^x $ for x positive is between -1 and 1 (excluding -1 and 1 except at x=0). Wait, for x positive, the absolute value is $ (\frac{5}{8})^x $, which is less than 1. Therefore, $ (-\frac{5}{8})^x $ is between -1 and 1. Therefore, the floor function can only be -1, 0, or 1. Therefore, only three values. Therefore, the answer is 3. Therefore, the number of distinct values is 3.

But wait, let me check for x=0. Is x=0 included? Yes. So f(0)=1. Then for x=1, f(x)=-1. For x=2, 0. For x=3, -1. For x=4, 0. So the range is { -1, 0, 1 }, so three distinct values. Therefore, the answer is 3. Therefore, the answer is \boxed{3}?

Wait, but let me check again. Maybe there's a mistake here. Let me think again. Suppose x is a non-integer. Wait, but earlier I thought that the function is only defined for integer x. But the problem says "defined for all values of x in [0, ∞) such that f(x) is a real number". So if x is not an integer, then $ (-\frac{5}{8})^x $ is not real. Therefore, the domain is only integers. Therefore, the function is defined only for integer x. Therefore, the range is { -1, 0, 1 }, so three values. Therefore, answer is 3. But maybe I need to check if there are other integer values?

Wait, for x=0, 1, 2, 3,... So for x=0, 1, 2,..., the function takes values 1, -1, 0, -1, 0,... So only three distinct values. Therefore, answer is 3. Therefore, I think the answer is 3. But let me check again.

Alternatively, maybe the problem is considering x as real numbers, and using some complex analysis? But then floor function is not defined. Therefore, the problem must be considering integer exponents. Therefore, answer is 3. Therefore, I think the answer is 3. However, let me check once more.

Wait, but maybe there's a case where for some non-integer x, the expression is real? For example, if x is a rational number with odd denominator? Let me think. Suppose x is a rational number, say x = p/q, where p and q are integers, q > 0, and p and q coprime. Then, $ (-\frac{5}{8})^{p/q} $ is real if and only if the denominator q is odd. Because if q is odd, then you can take the q-th root of a negative number. For example, $ (-\frac{5}{8})^{1/3} $ is real. However, if q is even, then it's not real. Therefore, if x is a rational number with odd denominator, then the expression is real. However, if x is irrational, then it's not real. Therefore, the domain of the function is all real numbers x in [0, ∞) such that x can be expressed as a rational number with odd denominator. However, this is a very complicated set, and the problem says "defined for all values of x in [0, ∞) such that f(x) is a real number". Therefore, the domain is all x in [0, ∞) for which $ (-\frac{5}{8})^x $ is real. Which includes integers (since they can be written as p/1, denominator 1, which is odd), and other rationals with odd denominators, and irrationals? No, irrationals can't be expressed as p/q with integer p, q. Therefore, the domain is all real numbers x in [0, ∞) that are rational numbers with odd denominators when expressed in lowest terms. However, this is a complicated domain. Therefore, the problem is not straightforward. Therefore, maybe the problem is intended to consider only integer exponents? Because otherwise, the function is defined on a complicated set, and the range would be more complex.

But the problem says "defined for all values of x in [0, ∞) such that f(x) is a real number". Therefore, the domain is all x in [0, ∞) where $ (-\frac{5}{8})^x $ is real. Therefore, as discussed, this includes integers and some other rationals. However, if we consider that, then the function is defined on a dense set in [0, ∞), but the problem is asking for the number of distinct values in the range. Therefore, maybe we need to analyze the function over all real numbers x where it is defined.

But this seems complicated. However, maybe the problem is intended to consider only integer exponents. Because otherwise, the problem becomes more complex. Let me check the original problem again. It says "defined for all values of x in [0, ∞) such that f(x) is a real number". Therefore, the function is defined on the set of x in [0, ∞) where $ (-\frac{5}{8})^x $ is real. Therefore, as discussed, this is the set of x which are rational numbers with odd denominators. However, even so, the function's range would depend on the values of x in that domain.

But this seems too complicated for a problem that is likely expecting an answer of 3. Therefore, maybe the problem is intended to consider x as integers. Therefore, given that, the answer is 3. However, to be thorough, let me consider the case where x is allowed to be any real number where $ (-\frac{5}{8})^x $ is real. Let me think about how the function behaves.

First, for x rational with odd denominator, say x = p/q, where q is odd. Then, $ (-\frac{5}{8})^{p/q} = \left( (-\frac{5}{8})^{1/q} \right)^p $. Since q is odd, $ (-\frac{5}{8})^{1/q} $ is real. Let me denote $ a = (-\frac{5}{8})^{1/q} $. Then, $ a $ is a real number. Since $ -\frac{5}{8} $ is negative, and q is odd, $ a $ is negative. Therefore, $ a = -|a| $, where $ |a| = (\frac{5}{8})^{1/q} $. Therefore, $ (-\frac{5}{8})^{p/q} = (-1)^p \cdot (\frac{5}{8})^{p/q} $. Therefore, depending on whether p is even or odd, the value is positive or negative. Therefore, for x rational with odd denominator, the value $ (-\frac{5}{8})^x $ is either positive or negative, depending on the parity of the numerator p. However, since x can be any such rational number, the value $ (-\frac{5}{8})^x $ can be positive or negative, and its magnitude is $ (\frac{5}{8})^x $, which is between 0 and 1 for x > 0. Therefore, similar to the integer case, for x > 0, the value is between -1 and 1, excluding -1 and 1. Therefore, the floor function would be -1, 0, or 1. However, for x=0, it's 1. Therefore, even if we consider x as real numbers where the expression is defined, the floor function would still only take values -1, 0, 1. Therefore, the range is still { -1, 0, 1 }, so three distinct values. Therefore, regardless of whether x is integer or rational with odd denominator, the range is still three values. Therefore, the answer is 3.

But wait, let me check for some specific non-integer x. For example, take x = 1/3. Then $ (-\frac{5}{8})^{1/3} = - (\frac{5}{8})^{1/3} \approx -0.855 $. Therefore, floor of that is -1. Similarly, x = 2/3: $ (-\frac{5}{8})^{2/3} = [(-\frac{5}{8})^{1/3}]^2 = [ - (\frac{5}{8})^{1/3} ]^2 = (\frac{5}{8})^{2/3} \approx 0.63 $. Therefore, floor is 0. Similarly, x = 1/1 (which is integer) gives -5/8 ≈ -0.625, floor -1. x = 2/1 = 2, gives 25/64 ≈ 0.39, floor 0. So even for non-integer x, the floor function still gives -1, 0, or 1. Therefore, regardless of x being integer or non-integer (as long as the expression is real), the floor function only takes those three values. Therefore, the range is { -1, 0, 1 }, so three distinct values. Therefore, the answer is 3.

But wait, let me check for x approaching infinity. As x approaches infinity, $ (-\frac{5}{8})^x $ approaches 0, since $ |\frac{5}{8}| < 1 $. Therefore, for x approaching infinity, the value approaches 0. Therefore, for x large enough, the value is between -1 and 1, but very close to 0. Therefore, floor is -1 or 0 depending on the sign. Therefore, even for non-integer x, the floor function is still -1 or 0. Therefore, the range is still { -1, 0, 1 }.

Therefore, regardless of the domain (as long as it's the set of x where the expression is real), the range has three elements. Therefore, the answer is 3. Therefore, the number of distinct values in the range of f(x) is 3. Therefore, the answer is \boxed{3}.

But wait, let me check once more. Suppose x is a non-integer rational with odd denominator. For example, x = 1/3, which we did. The value is approximately -0.855, floor is -1. If x is 1/5, then $ (-\frac{5}{8})^{1/5} \approx - (0.625)^{1/5} \approx -0.92 $, floor is -1. If x is 1/1000, then $ (-\frac{5}{8})^{1/1000} $ is a negative number very close to -1, since raising to a small exponent. Wait, no. Wait, $ (-\frac{5}{8})^{1/n} $ as n approaches infinity. Let me think. For x approaching 0 from the positive side, $ (-\frac{5}{8})^x $. Wait, if x approaches 0, then $ (-\frac{5}{8})^x $ approaches 1. For example, x approaching 0, even if x is a non-integer. For example, x approaching 0 from the positive side, say x = 0.1. Then $ (-\frac{5}{8})^{0.1} $. Wait, but this is not real. Wait, if x is 0.1, which is not rational with odd denominator? Wait, 0.1 is 1/10, which has denominator 10, which is even. Therefore, $ (-\frac{5}{8})^{0.1} $ is not real. Therefore, x=0.1 is not in the domain. Therefore, the domain is only those x which are rational numbers with odd denominators. Therefore, for example, x=1/3, 1/5, 1/7, etc. So, for x approaching 0 from the right, with x being rational with odd denominator, say x = 1/n where n is odd. Then, $ (-\frac{5}{8})^{1/n} $ is a negative number approaching -1 as n approaches infinity? Wait, no. Wait, as n approaches infinity, $ (-\frac{5}{8})^{1/n} $ approaches 1? Wait, no. Let me think. Let me take the limit as n approaches infinity of $ (-\frac{5}{8})^{1/n} $. Let me write it as $ e^{\frac{1}{n} \ln(-\frac{5}{8})} $. But ln(-5/8) is not real. Therefore, this is not defined. Wait, but earlier we said that for x rational with odd denominator, the expression is real. So for x = 1/n, with n odd, then $ (-\frac{5}{8})^{1/n} $ is real. Let me compute the limit as n approaches infinity of $ (-\frac{5}{8})^{1/n} $. Let me write it as $ (-1)^{1/n} \cdot (\frac{5}{8})^{1/n} $. Since n is odd, $ (-1)^{1/n} = -1 $. Therefore, the expression is $ - (\frac{5}{8})^{1/n} $. As n approaches infinity, $ (\frac{5}{8})^{1/n} $ approaches 1, so the whole expression approaches -1. Therefore, as x approaches 0 from the right (with x being rational with odd denominator), the value approaches -1. Therefore, for x approaching 0 from the right, the value approaches -1. Therefore, for x very close to 0, the value is slightly greater than -1? Wait, no. Wait, if x approaches 0 from the right, then $ (-\frac{5}{8})^x $ approaches 1? Wait, no. Wait, if x approaches 0, then for x positive, $ (-\frac{5}{8})^x $ is... Wait, if x is approaching 0, but x is rational with odd denominator. For example, take x = 1/n, n odd, approaching 0. Then, as n increases, x approaches 0. Then, $ (-\frac{5}{8})^{1/n} = - (\frac{5}{8})^{1/n} $. As n approaches infinity, $ (\frac{5}{8})^{1/n} $ approaches 1, so the whole expression approaches -1. Therefore, for x approaching 0 from the right, the value approaches -1. Therefore, for x very small positive, the value is slightly greater than -1? Wait, no. Wait, $ (\frac{5}{8})^{1/n} $ is less than 1 for n > 0. Therefore, $ - (\frac{5}{8})^{1/n} $ is greater than -1. Therefore, as x approaches 0 from the right, the value approaches -1 from above. Therefore, for x approaching 0, the value is between -1 and 0, approaching -1. Therefore, floor of that is -1. Therefore, even for x approaching 0, floor is -1. Except when x=0, which is 1. Therefore, the floor function is -1 for all x > 0, except when the value is between 0 and 1, which occurs when x is even? Wait, no. Wait, for x being rational with odd denominator, the value is negative or positive depending on whether the exponent is even or odd? Wait, earlier we saw that $ (-\frac{5}{8})^x = (-1)^p \cdot (\frac{5}{8})^x $, where x = p/q with q odd. Wait, no, earlier we had x = p/q, then $ (-\frac{5}{8})^{p/q} = (-1)^p \cdot (\frac{5}{8})^{p/q} $. Therefore, if p is even, the value is positive; if p is odd, the value is negative. Therefore, for x rational with odd denominator, depending on the numerator's parity, the value is positive or negative. Therefore, for example, if x is 2/3 (p=2, q=3), then the value is positive. If x is 1/3, it's negative. Therefore, for x rational with odd denominator, the value can be positive or negative. Therefore, for positive x, depending on whether the exponent is even or odd (in terms of numerator), the value is positive or negative. Therefore, for x positive, the value is between -1 and 1, and floor is -1, 0, or 1. However, when is the value between 0 and 1? When the exponent is even (in terms of numerator). For example, x = 2/3, which is even numerator, gives positive value between 0 and 1. Similarly, x = 4/5, which is even numerator, gives positive value between 0 and 1. Therefore, floor is 0. For x with odd numerator, the value is negative between -1 and 0, floor is -1. For x=0, floor is 1. Therefore, the range is still { -1, 0, 1 }, so three distinct values. Therefore, regardless of x being integer or non-integer (as long as the expression is real), the range is three values. Therefore, the answer is 3.

Therefore, I think the answer is 3. Therefore, the number of distinct values in the range of f(x) is \boxed{3}.

**Final Answer**
\boxed{3}

To determine how many **distinct values** exist in the range of the function
$$
f(x) = \left\lfloor \left(-\frac{5}{8}\right)^x \right\rfloor
$$
for $ x \in [0, \infty) $, we must first understand the behavior of the expression inside the floor function:
$$
\left(-\frac{5}{8}\right)^x
$$

---

### Step 1: Understanding the Domain

The base $ -\frac{5}{8} $ is a **negative real number** with absolute value less than 1. When raising a negative number to a real exponent, the result is **not always real**. Specifically:

- If $ x $ is an **integer**, then $ \left(-\frac{5}{8}\right)^x $ is real.
- If $ x $ is a **non-integer rational number** with an **odd denominator**, then the expression is also real (since we can take an odd root of a negative number).
- If $ x $ is an **irrational number**, the expression is **not real**.

However, the problem states that the function is defined for all $ x \in [0, \infty) $ such that $ f(x) $ is a real number. Therefore, the domain of $ f(x) $ is the set of all $ x \in [0, \infty) $ for which $ \left(-\frac{5}{8}\right)^x $ is real. This includes:

- All **non-negative integers** $ x = 0, 1, 2, \dots $
- All **non-integer rational numbers** with **odd denominators** in their reduced form.

But for the purpose of determining the **range**, we can consider the behavior of the function over the entire domain where it is defined.

---

### Step 2: Behavior of the Function

Let’s analyze the expression $ \left(-\frac{5}{8}\right)^x $ for different values of $ x $:

- **When $ x = 0 $:**
  $$
  \left(-\frac{5}{8}\right)^0 = 1 \Rightarrow f(0) = \lfloor 1 \rfloor = 1
  $$

- **When $ x $ is a **positive integer** (odd or even):**
  - If $ x $ is **odd**, then $ \left(-\frac{5}{8}\right)^x $ is **negative** and lies in the interval $ (-1, 0) $, so
    $$
    f(x) = \lfloor \text{negative number in } (-1, 0) \rfloor = -1
    $$
  - If $ x $ is **even**, then $ \left(-\frac{5}{8}\right)^x $ is **positive** and lies in the interval $ (0, 1) $, so
    $$
    f(x) = \lfloor \text{positive number in } (0, 1) \rfloor = 0
    $$

- **When $ x $ is a **non-integer rational number** with an **odd denominator** (e.g., $ x = \frac{1}{3}, \frac{2}{5}, \dots $):**
  - If the numerator is **odd**, the result is **negative** and in $ (-1, 0) $, so $ f(x) = -1 $
  - If the numerator is **even**, the result is **positive** and in $ (0, 1) $, so $ f(x) = 0 $

- **As $ x \to \infty $:**
  The magnitude of $ \left(-\frac{5}{8}\right)^x $ approaches 0, so the value is always in $ (-1, 1) $, and the floor function still yields either $ -1 $, $ 0 $, or $ 1 $.

---

### Step 3: Determining the Range

From the above, we observe that the function $ f(x) $ can only take the following **three distinct values**:

- $ f(x) = 1 $ when $ x = 0 $
- $ f(x) = -1 $ when $ x $ is a **positive rational number with odd numerator** (i.e., odd exponent)
- $ f(x) = 0 $ when $ x $ is a **positive rational number with even numerator** (i.e., even exponent)

Thus, the **range** of $ f(x) $ is:
$$
\{ -1, 0, 1 \}
$$

---

### Final Answer

$$
\boxed{3}
$$

Solution: Since $-\frac58$ is a negative number, $f(x)$ is only defined for integer values of $x$, and will alternate between positive and negative values. Additionally, $\left|-\frac58\right|< 1$, so $|f(x)|$ will continually decrease and approach 0 as $x$ increases in the interval $x\ge0$. Therefore, the largest positive value will occur at $x=0$, giving us the positive upper bound of $\left\lfloor\left(-\frac58\right)^0\right\rfloor=1$. The negative value that is greatest in magnitude then occurs at the next integer value of $x$: $x=1$, giving us the negative lower bound of $\left\lfloor\left(-\frac58\right)^1\right\rfloor=-1$. This tells us that $-1 \le f(x) \le 1$. Since the $f(x)$ must be an integer, the only possible distinct values contained in the range are -1, 0, and 1. This gives us a total of $\boxed{3}$ values of $f(x)$ when $x\ge0$.