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).

Let me write down these equations. Let me denote R(x) = mx + n. Then:

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

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

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

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

Also, we know that P(0) = 5. Since P(x) is monic cubic, let me write P(x) as x³ + ax² + bx + c. Then P(0) = c = 5. So c = 5.

Therefore, P(x) = x³ + ax² + bx + 5.

Now, let me compute P(1), P(2), P(3), P(4) in terms of a, b, and then set up equations.

First, P(1) = 1³ + a(1)² + b(1) + 5 = 1 + a + b + 5 = a + b + 6

Similarly, 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

But from earlier, 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

Therefore, we can set up equations:

1) a + b + 6 = m + n

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

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

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

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

Equation 1: a + b + 6 = m + n

Equation 2: 4a + 2b + 13 = 4m + 2n

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

Equation 4: 16a + 4b + 69 = 4m + n

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

First, maybe I can express equations 2, 3, 4 in terms of m and n, and then relate them to equations 1.

Alternatively, since equations 2, 3, 4 involve m and n, perhaps I can solve for m and n first?

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

Let me note that equations 2 and 3 both have 2n. Let me subtract equation 2 from equation 3:

Equation 3 - Equation 2:

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

Left side: 5a + b + 19

Right side: 2m

So, 5a + b + 19 = 2m. Let me call this Equation 5.

Similarly, let me subtract equation 2 from equation 4? Wait, equation 4 is 16a + 4b + 69 = 4m + n. Let me see.

Alternatively, maybe first handle equations 2 and 4. Let me see.

Alternatively, let me consider equations 2 and 4. Let me write equations 2 and 4:

Equation 2: 4a + 2b + 13 = 4m + 2n

Equation 4: 16a + 4b + 69 = 4m + n

Let me denote equation 2 as:

Equation 2: 4a + 2b + 13 = 4m + 2n

Equation 4: 16a + 4b + 69 = 4m + n

Let me subtract equation 2 from equation 4:

(16a + 4b + 69) - (4a + 2b + 13) = (4m + n) - (4m + 2n)

Left side: 12a + 2b + 56

Right side: -n

Therefore:

12a + 2b + 56 = -n

Let me call this Equation 6: 12a + 2b + 56 = -n

Similarly, let me look at equation 1: a + b + 6 = m + n. Let me call this Equation 1.

So, if I can express m and n in terms of a and b, maybe?

Alternatively, let me try to find another equation. Let me also consider equation 5: 5a + b + 19 = 2m.

So, if I can express m from equation 5: m = (5a + b + 19)/2.

Similarly, from equation 1: m + n = a + b + 6 => n = a + b + 6 - m.

So, substituting m from equation 5 into this:

n = a + b + 6 - (5a + b + 19)/2

Let me compute that:

Multiply numerator:

n = [2(a + b + 6) - 5a - b - 19]/2

Compute numerator:

2a + 2b + 12 - 5a - b - 19 = (-3a + b - 7)

Therefore, n = (-3a + b - 7)/2. Let me call this Equation 7.

Now, Equation 6 is 12a + 2b + 56 = -n. So, substituting n from Equation 7:

12a + 2b + 56 = - [ (-3a + b - 7)/2 ]

Multiply both sides by 2 to eliminate denominator:

2*(12a + 2b + 56) = 3a - b + 7

Left side: 24a + 4b + 112

So:

24a + 4b + 112 = 3a - b + 7

Bring all terms to left side:

24a + 4b + 112 - 3a + b - 7 = 0

Simplify:

21a + 5b + 105 = 0

So, 21a + 5b = -105. Let me call this Equation 8.

Now, let me see if I can find another equation involving a and b. Let me check equation 5: 5a + b + 19 = 2m. But we also have equation 1 and equation 7. Wait, maybe I can use equation 3?

Wait, equation 3 was 9a + 3b + 32 = 6m + 2n. Let me try to use that. Let me express m and n in terms of a and b.

From equation 5: m = (5a + b + 19)/2

From equation 7: n = (-3a + b - 7)/2

So, substitute m and n into equation 3:

Left side: 9a + 3b + 32

Right side: 6m + 2n = 6*(5a + b + 19)/2 + 2*(-3a + b - 7)/2

Simplify:

= 3*(5a + b + 19) + (-3a + b - 7)

= 15a + 3b + 57 - 3a + b - 7

= 12a + 4b + 50

Therefore, equation 3 becomes:

9a + 3b + 32 = 12a + 4b + 50

Bring all terms to left side:

9a + 3b + 32 - 12a - 4b - 50 = 0

Simplify:

-3a - b - 18 = 0 => -3a - b = 18 => 3a + b = -18. Let me call this Equation 9.

Now, we have Equation 8: 21a + 5b = -105 and Equation 9: 3a + b = -18.

So, we can solve these two equations. Let me solve Equation 9 for b: b = -18 - 3a. Then substitute into Equation 8:

21a + 5*(-18 - 3a) = -105

Compute:

21a - 90 - 15a = -105

Simplify:

6a - 90 = -105

Add 90 to both sides:

6a = -15 => a = -15/6 = -5/2.

So, a = -5/2. Then from Equation 9: b = -18 - 3*(-5/2) = -18 + 15/2 = (-36/2 + 15/2) = (-21)/2 = -10.5 or -21/2.

So, a = -5/2, b = -21/2.

Now, let me check if these values satisfy Equation 8:

21a + 5b = 21*(-5/2) + 5*(-21/2) = (-105/2) + (-105/2) = -210/2 = -105. Yes, that's correct.

So, now we have a and b. Now, let me find m and n.

From equation 5: m = (5a + b + 19)/2.

Plugging in a = -5/2 and b = -21/2:

5a = 5*(-5/2) = -25/2

b = -21/2

So, 5a + b + 19 = (-25/2 -21/2) + 19 = (-46/2) + 19 = -23 + 19 = -4

Therefore, m = -4 / 2 = -2.

So, m = -2.

From equation 7: n = (-3a + b - 7)/2.

Compute:

-3a = -3*(-5/2) = 15/2

b = -21/2

So, -3a + b -7 = 15/2 -21/2 -7 = (-6/2) -7 = -3 -7 = -10

Therefore, n = -10 / 2 = -5.

So, n = -5.

Therefore, R(x) = mx + n = -2x -5.

Let me check if this is consistent with the earlier equations.

For example, check equation 1: a + b + 6 = m + n.

a + b + 6 = (-5/2) + (-21/2) + 6 = (-26/2) + 6 = -13 + 6 = -7

m + n = -2 + (-5) = -7. Correct.

Similarly, check equation 2: 4a + 2b + 13 = 4m + 2n.

Compute left side: 4*(-5/2) + 2*(-21/2) + 13 = (-10) + (-21) + 13 = -18

Right side: 4*(-2) + 2*(-5) = -8 -10 = -18. Correct.

Equation 3: 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. Correct.

Equation 4: 16a + 4b + 69 = 4m + n.

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 is:

P(x) = x³ - (5/2)x² - (21/2)x + 5.

But maybe I should write it with fractions or convert to integers? Let me check if I can write it as:

Multiply through by 2 to eliminate denominators? But since the problem asks for P(5), maybe I can compute it directly.

Alternatively, maybe I can write P(x) as:

P(x) = x³ - (5/2)x² - (21/2)x + 5.

But let me check if there's another way to find P(5) without having to compute the entire polynomial? Maybe using the remainder information?

Wait, but since we have the polynomial, maybe it's easier to compute P(5) directly. Let me do that.

Compute P(5):

P(5) = 5³ - (5/2)(5²) - (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 = 15.

Wait, 125 - 115 is 10, plus 5 is 15. So P(5) = 15?

But let me check my calculation again to be sure.

Alternatively, maybe I made a mistake in the calculation. Let me recompute:

P(5) = 5³ - (5/2)(5²) - (21/2)(5) + 5

First term: 125

Second term: (5/2)(25) = 125/2 = 62.5

Third term: (21/2)(5) = 105/2 = 52.5

Fourth term: 5

So:

125 - 62.5 - 52.5 + 5

Compute 125 - 62.5 = 62.5

Then 62.5 - 52.5 = 10

Then 10 + 5 = 15. Yes, same result. So P(5) = 15.

But let me check if there's another way to get this result, maybe using the remainders?

Alternatively, since we know R(x) = -2x -5, and we know that P(x) = (x - 1)(x - 4)Q(x) + R(x). Also, since P(x) is cubic, Q(x) is linear. Let me write Q(x) as 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 of (x - 1)(x - 4)(px + q) is x² * px = px³. Since P(x) is monic, the coefficient of x³ is 1, so p must be 1. Therefore, Q(x) = x + q.

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

Let me expand this expression and compare with the previous expression for P(x).

First, expand (x - 1)(x - 4):

(x - 1)(x - 4) = x² - 5x + 4

Multiply by (x + q):

(x² - 5x + 4)(x + q) = x³ + qx² -5x² -5qx + 4x + 4q

Simplify:

x³ + (q - 5)x² + (-5q + 4)x + 4q

Then subtract 2x + 5:

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

Combine like terms:

x³ + (q - 5)x² + [(-5q + 4) - 2]x + (4q -5)

Simplify:

x³ + (q - 5)x² + (-5q + 2)x + (4q -5)

But we also have from earlier that P(x) = x³ - (5/2)x² - (21/2)x + 5. Therefore, equate coefficients:

For x³: 1 = 1, which is 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:

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

For constant term: 4q -5 = 4*(5/2) -5 = 10 -5 = 5. Correct. Perfect.

Therefore, q = 5/2. Therefore, P(x) can be written as:

(x - 1)(x - 4)(x + 5/2) - 2x -5.

But since we already found P(5) = 15, maybe we can check this expression for P(5):

Compute P(5) = (5 - 1)(5 - 4)(5 + 5/2) - 2*5 -5

Compute each part:

(5 - 1) = 4

(5 - 4) = 1

(5 + 5/2) = 15/2

Multiply them: 4 * 1 * 15/2 = 4*(15/2) = 30

Then subtract 2*5 +5 = 10 +5 = 15. Therefore, P(5) = 30 - 15 = 15. Correct. So that's another way to compute it.

Therefore, the answer is 15. But let me check once more with the original polynomial.

Original polynomial: P(x) = x³ - (5/2)x² - (21/2)x + 5.

Compute P(5):

5³ = 125

(5/2)x² at x=5: (5/2)*(25) = 125/2

(21/2)x at x=5: (21/2)*5 = 105/2

So:

P(5) = 125 - 125/2 - 105/2 + 5 = 125 - (125 + 105)/2 + 5 = 125 - 230/2 + 5 = 125 - 115 + 5 = 15. Correct.

Therefore, the answer is 15. However, let me check if there's another way to get this result without computing the entire polynomial, maybe using the given conditions?

Alternatively, since we know that P(x) = (x - 1)(x - 4)Q(x) + R(x), and also P(x) = (x - 2)(x - 3)S(x) + 2R(x). Since we know R(x) = -2x -5, then maybe we can use these expressions?

But since we already have the answer, maybe that's not necessary. However, just to ensure that there are no mistakes in the process, let me check if P(0) = 5. From the polynomial:

P(0) = 0³ - (5/2)(0)² - (21/2)(0) + 5 = 5. Correct. So that's good.

Another check: Let me compute P(1). From the polynomial:

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

Similarly, P(4) = 64 - (5/2)(16) - (21/2)(4) +5 = 64 - 40 - 42 +5 = 64 - 82 +5 = -13. Which should be R(4) = -2(4) -5 = -8 -5 = -13. Correct.

Similarly, P(2) = 8 - (5/2)(4) - (21/2)(2) +5 = 8 - 10 -21 +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 - 45/2 -63/2 +5 = 27 +5 - (45 +63)/2 = 32 - 108/2 = 32 -54 = -22. Which should be 2R(3) = 2*(-2*3 -5) = 2*(-6 -5) = 2*(-11) = -22. Correct.

All checks out. Therefore, P(5) = 15 is correct.

But just to make sure, let me think if there's another approach. For example, using the Chinese Remainder Theorem or something else?

Alternatively, since 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. Since P(x) is cubic, A(x) and B(x) are linear.

But since P(x) is monic, then A(x) must be monic? Let me see. If P(x) - R(x) is (x - 1)(x - 4)A(x), then the leading term of P(x) is x³, and the leading term of (x - 1)(x - 4)A(x) is x² * A(x)'s leading term. Since P(x) is monic, then A(x) must be monic. Similarly for B(x). Therefore, A(x) = x + c, B(x) = x + d.

But maybe this approach is similar to what I did before. However, since we already have the answer, maybe it's redundant.

Alternatively, since we know R(x) = -2x -5, then P(x) = (x - 1)(x - 4)Q(x) -2x -5. Also, since P(x) is monic cubic, Q(x) is linear, as before. So, if I can find Q(x), but we already did that.

Alternatively, since we have P(5), maybe we can use the fact that P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, P(5) - R(5) is divisible by (5 - 1)(5 - 4) = 4*1 = 4. Similarly, P(5) - 2R(5) is divisible by (5 - 2)(5 - 3) = 3*2 = 6. But since we already know P(5) = 15, let me check:

R(5) = -2*5 -5 = -15. Therefore, P(5) - R(5) = 15 - (-15) = 30, which is divisible by 4? 30 divided by 4 is 7.5, which is not integer. Wait, but that contradicts? Wait, but P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, P(5) - R(5) should be divisible by (5 - 1)(5 - 4) = 4. But 30 is not divisible by 4. Wait, that seems contradictory. What's wrong here?

Wait, hold on. Wait, if P(x) - R(x) is divisible by (x - 1)(x - 4), then P(x) - R(x) = (x - 1)(x - 4)Q(x). Therefore, for any x, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, in particular, when x = 5, P(5) - R(5) is divisible by (5 - 1)(5 - 4) = 4. But according to our calculation, P(5) - R(5) = 15 - (-15) = 30. 30 divided by 4 is 7.5, which is not integer. That seems contradictory. Wait, but this suggests that there is a mistake in my previous reasoning?

Wait, but earlier checks showed that P(x) is correct. So why is this discrepancy?

Wait, maybe my mistake is in assuming that P(x) - R(x) is divisible by (x - 1)(x - 4). Wait, the problem says that when P(x) is divided by (x - 1)(x - 4), the remainder is R(x). Therefore, P(x) = (x - 1)(x - 4)Q(x) + R(x). Therefore, P(x) - R(x) = (x - 1)(x - 4)Q(x), which is divisible by (x - 1)(x - 4). Therefore, for any x, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, in particular, when x = 5, P(5) - R(5) must be divisible by (5 - 1)(5 - 4) = 4. However, according to our calculation, P(5) - R(5) = 15 - (-15) = 30, which is not divisible by 4. Therefore, there must be a mistake here. Wait, but earlier checks showed that P(x) is correct. So where is the error?

Wait, let me check R(x). Earlier, we found R(x) = -2x -5. Let me check if that is correct. From the equations, we had:

P(1) = -7, which is R(1) = -2(1) -5 = -7. Correct.

Similarly, P(4) = -13, which is R(4) = -2(4) -5 = -13. Correct.

Therefore, R(x) is correct. Then P(5) - R(5) = 15 - (-15) = 30. But 30 divided by 4 is 7.5. But according to the division algorithm, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, P(5) - R(5) must be divisible by (5 - 1)(5 - 4) = 4. But 30 is not divisible by 4. Therefore, there is a contradiction here. Which means that there is a mistake in my previous calculations. Wait, but all the checks for P(1), P(2), P(3), P(4) were correct. How is this possible?

Wait, let me check P(5) again. If P(x) = x³ - (5/2)x² - (21/2)x + 5, then P(5) is:

5³ = 125

(5/2)x² at x=5: (5/2)(25) = 125/2 = 62.5

(21/2)x at x=5: (21/2)(5) = 105/2 = 52.5

So P(5) = 125 - 62.5 - 52.5 + 5 = 125 - 115 + 5 = 15. Correct.

R(5) = -2*5 -5 = -15. Therefore, P(5) - R(5) = 15 - (-15) = 30. But 30 divided by 4 is 7.5. But according to the division algorithm, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, P(5) - R(5) should be divisible by (5 - 1)(5 - 4) = 4. But 30 is not divisible by 4. Therefore, there is a contradiction. Which means that there is a mistake in my previous steps. But where?

Wait, let me check the equations again. Let me check the equations for P(1), P(2), P(3), P(4). Earlier, I had:

From the polynomial P(x) = x³ - (5/2)x² - (21/2)x + 5.

Compute P(1):

1 - (5/2) - (21/2) + 5 = 1 - 26/2 + 5 = 1 -13 +5 = -7. Correct.

P(2):

8 - (5/2)(4) - (21/2)(2) +5 = 8 - 10 -21 +5 = (8 +5) - (10 +21) = 13 - 31 = -18. Which is 2R(2) = 2*(-2*2 -5) = 2*(-9) = -18. Correct.

P(3):

27 - (5/2)(9) - (21/2)(3) +5 = 27 - 45/2 -63/2 +5 = 27 +5 - (45 +63)/2 = 32 - 108/2 = 32 -54 = -22. Which is 2R(3) = 2*(-2*3 -5) = 2*(-11) = -22. Correct.

P(4):

64 - (5/2)(16) - (21/2)(4) +5 = 64 - 40 -42 +5 = 64 - 82 +5 = -13. Which is R(4) = -2*4 -5 = -13. Correct.

So all these are correct. Then why is P(5) - R(5) not divisible by 4?

Wait, maybe my mistake is in the assumption that P(x) - R(x) is divisible by (x - 1)(x - 4). Wait, the division algorithm says that when you divide P(x) by (x - 1)(x - 4), you get quotient Q(x) and remainder R(x). Therefore, P(x) = (x - 1)(x - 4)Q(x) + R(x). Therefore, P(x) - R(x) = (x - 1)(x - 4)Q(x). Therefore, for any x, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, in particular, when x = 5, P(5) - R(5) must be divisible by (5 - 1)(5 - 4) = 4. But according to our calculation, P(5) - R(5) = 30, which is not divisible by 4. Therefore, there is a contradiction. Therefore, my previous solution must have an error.

Wait, but how? All the checks for P(1), P(2), P(3), P(4) are correct. Let me check the value of Q(x). Earlier, I said that Q(x) = x + q, with q = 5/2. Therefore, P(x) = (x - 1)(x - 4)(x + 5/2) - 2x -5. Let me compute P(5) using this expression:

(x - 1)(x - 4)(x + 5/2) - 2x -5 at x = 5:

(5 - 1)(5 - 4)(5 + 5/2) - 2*5 -5 = 4*1*(15/2) - 10 -5 = 30 - 15 = 15. Correct. So P(5) = 15. Then R(5) = -2*5 -5 = -15. Therefore, P(5) - R(5) = 15 - (-15) = 30. But according to the division algorithm, this should be divisible by (5 - 1)(5 - 4) = 4. But 30 is not divisible by 4. What is happening here?

Wait, maybe my mistake is in the definition of R(x). Wait, the problem says that when P(x) is divided by (x - 1)(x - 4), the remainder is R(x). But R(x) is a linear polynomial. However, when we divide P(x) by (x - 1)(x - 4), the remainder is R(x), which is of degree less than 2, so degree 1 or 0. But in our case, R(x) is linear. However, when we compute P(x) - R(x), it is divisible by (x - 1)(x - 4). Therefore, P(x) - R(x) = (x - 1)(x - 4)Q(x). Therefore, for any x, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, in particular, when x = 5, P(5) - R(5) must be divisible by (5 - 1)(5 - 4) = 4. But 30 is not divisible by 4. Therefore, there is a contradiction. Therefore, my previous solution is wrong. But how?

Wait, but all the checks for P(1), P(2), P(3), P(4) are correct. Let me check if P(x) - R(x) is divisible by (x - 1)(x - 4). Let me compute P(x) - R(x):

P(x) - R(x) = x³ - (5/2)x² - (21/2)x + 5 - (-2x -5) = x³ - (5/2)x² - (21/2)x + 5 + 2x + 5 = x³ - (5/2)x² - (17/2)x + 10.

Now, let me check if this polynomial is divisible by (x - 1)(x - 4) = x² - 5x + 4.

Let me perform polynomial division or factor it.

Let me try to factor x³ - (5/2)x² - (17/2)x + 10.

Alternatively, since we know that (x - 1)(x - 4) is a factor, then it should be divisible by x² - 5x + 4. Let me perform the division.

Divide x³ - (5/2)x² - (17/2)x + 10 by x² - 5x + 4.

Let me write it as:

Dividend: x³ - (5/2)x² - (17/2)x + 10

Divisor: x² - 5x + 4

First term: x³ / x² = x. Multiply divisor by x: x³ - 5x² + 4x.

Subtract this from dividend:

(x³ - (5/2)x² - (17/2)x + 10) - (x³ - 5x² + 4x) = [0x³] + [(-5/2 + 5)x²] + [(-17/2 - 4)x] + 10

Compute coefficients:

-5/2 + 5 = -5/2 + 10/2 = 5/2

-17/2 - 4 = -17/2 - 8/2 = -25/2

So, the remainder is (5/2)x² - (25/2)x + 10.

Now, divide this remainder by the divisor x² - 5x + 4.

Next term: (5/2)x² / x² = 5/2. Multiply divisor by 5/2: (5/2)x² - (25/2)x + 10.

Subtract this from the remainder:

(5/2)x² - (25/2)x + 10 - [(5/2)x² - (25/2)x + 10] = 0. Perfect, no remainder.

Therefore, P(x) - R(x) = (x - 1)(x - 4)(x + 5/2). Therefore, it is divisible by (x - 1)(x - 4). Therefore, there is no contradiction. So why is P(5) - R(5) = 30 not divisible by 4?

Wait, because P(x) - R(x) is divisible by (x - 1)(x - 4), which means that for any x, P(x) - R(x) is divisible by (x - 1)(x - 4). But when x = 5, (x - 1)(x - 4) = 4*1 = 4. Therefore, P(5) - R(5) must be divisible by 4. But 30 is not divisible by 4. Wait, but according to the division, P(x) - R(x) = (x - 1)(x - 4)(x + 5/2). Therefore, P(5) - R(5) = (5 - 1)(5 - 4)(5 + 5/2) = 4*1*(15/2) = 30. Which is equal to 30. But 30 is divisible by 4? No, 30 divided by 4 is 7.5. Wait, but according to the polynomial division, P(x) - R(x) is divisible by (x - 1)(x - 4), so P(x) - R(x) = (x - 1)(x - 4)Q(x). Therefore, when x = 5, P(5) - R(5) = (5 - 1)(5 - 4)Q(5) = 4*1*Q(5). Therefore, P(5) - R(5) = 4*Q(5). Therefore, it must be divisible by 4. But according to our calculation, P(5) - R(5) = 30, which would mean that Q(5) = 30 / 4 = 7.5. But Q(x) is a linear polynomial, which is x + 5/2. Therefore, Q(5) = 5 + 5/2 = 15/2 = 7.5. Which is correct. Therefore, 4 * 7.5 = 30. So, 30 is divisible by 4? No, but 30 is equal to 4 * 7.5. But 7.5 is not an integer. However, the division algorithm doesn't require Q(x) to have integer coefficients. Q(x) is a polynomial with real coefficients, not necessarily integer. Therefore, there is no contradiction. My mistake was assuming that Q(x) must have integer coefficients, but it doesn't. Therefore, everything is correct. So P(5) = 15 is correct.

Therefore, the answer is 15. So I think my initial solution is correct, and the confusion was due to my incorrect assumption about Q(x) having integer coefficients. Therefore, the final answer is 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: General Form of the Polynomial

Since $ P(x) $ is a monic cubic polynomial, we can write it as:

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

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

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

Thus, the polynomial becomes:

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

---

### Step 2: Use the Remainder Theorem

Let $ R(x) = mx + n $, a linear polynomial (since the remainder when dividing by a quadratic is of degree less than 2).

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) $:

- $ 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 $

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

1. $ a + b + 6 = m + n $
2. $ 4a + 2b + 13 = 4m + 2n $
3. $ 9a + 3b + 32 = 6m + 2n $
4. $ 16a + 4b + 69 = 4m + n $

---

### Step 3: Solve the System of Equations

From equations (1) and (2), subtract (1) from (2):

$$
(4a + 2b + 13) - (a + b + 6) = (4m + 2n) - (m + n)
\Rightarrow 3a + b + 7 = 3m + n \tag{5}
$$

From equations (2) and (4), subtract (2) from (4):

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

From equation (5), solve for $ n $:

$$
n = 3a + b + 7 - 3m \tag{7}
$$

Substitute (7) into (6):

$$
12a + 2b + 56 = - (3a + b + 7 - 3m)
\Rightarrow 12a + 2b + 56 = -3a - b - 7 + 3m
\Rightarrow 15a + 3b + 63 = 3m \tag{8}
$$

Now, from equation (1):

$$
a + b + 6 = m + n \Rightarrow n = a + b + 6 - m \tag{9}
$$

Substitute (9) into (6):

$$
12a + 2b + 56 = - (a + b + 6 - m)
\Rightarrow 12a + 2b + 56 = -a - b - 6 + m
\Rightarrow 13a + 3b + 62 = m \tag{10}
$$

Now, from (8) and (10), equate expressions for $ m $:

$$
15a + 3b + 63 = 13a + 3b + 62
\Rightarrow 2a = -1 \Rightarrow a = -\frac{1}{2}
$$

Substitute $ a = -\frac{1}{2} $ into (10):

$$
13(-\frac{1}{2}) + 3b + 62 = m
\Rightarrow -\frac{13}{2} + 3b + 62 = m
\Rightarrow 3b + \frac{101}{2} = m
$$

Now substitute into (1):

$$
a + b + 6 = m + n
\Rightarrow -\frac{1}{2} + b + 6 = m + n
\Rightarrow b + \frac{11}{2} = m + n
$$

From (9): $ n = a + b + 6 - m = -\frac{1}{2} + b + 6 - m = b + \frac{11}{2} - m $

So:

$$
b + \frac{11}{2} = m + (b + \frac{11}{2} - m) \Rightarrow \text{True}
$$

Now, from (10): $ m = 3b + \frac{101}{2} $

Substitute into (1):

$$
a + b + 6 = m + n = m + (b + \frac{11}{2} - m) = b + \frac{11}{2}
\Rightarrow -\frac{1}{2} + b + 6 = b + \frac{11}{2}
\Rightarrow \frac{11}{2} = \frac{11}{2}
$$

So the system is consistent.

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[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).

Let me write down these equations. Let me denote R(x) = mx + n. Then:

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

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

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

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

Also, we know that P(0) = 5. Since P(x) is monic cubic, let me write P(x) as x³ + ax² + bx + c. Then P(0) = c = 5. So c = 5.

Therefore, P(x) = x³ + ax² + bx + 5.

Now, let me compute P(1), P(2), P(3), P(4) in terms of a, b, and then set up equations.

First, P(1) = 1³ + a(1)² + b(1) + 5 = 1 + a + b + 5 = a + b + 6

Similarly, 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

But from earlier, 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

Therefore, we can set up equations:

1) a + b + 6 = m + n

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

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

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

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

Equation 1: a + b + 6 = m + n

Equation 2: 4a + 2b + 13 = 4m + 2n

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

Equation 4: 16a + 4b + 69 = 4m + n

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

First, maybe I can express equations 2, 3, 4 in terms of m and n, and then relate them to equations 1.

Alternatively, since equations 2, 3, 4 involve m and n, perhaps I can solve for m and n first?

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

Let me note that equations 2 and 3 both have 2n. Let me subtract equation 2 from equation 3:

Equation 3 - Equation 2:

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

Left side: 5a + b + 19

Right side: 2m

So, 5a + b + 19 = 2m. Let me call this Equation 5.

Similarly, let me subtract equation 2 from equation 4? Wait, equation 4 is 16a + 4b + 69 = 4m + n. Let me see.

Alternatively, maybe first handle equations 2 and 4. Let me see.

Alternatively, let me consider equations 2 and 4. Let me write equations 2 and 4:

Equation 2: 4a + 2b + 13 = 4m + 2n

Equation 4: 16a + 4b + 69 = 4m + n

Let me denote equation 2 as:

Equation 2: 4a + 2b + 13 = 4m + 2n

Equation 4: 16a + 4b + 69 = 4m + n

Let me subtract equation 2 from equation 4:

(16a + 4b + 69) - (4a + 2b + 13) = (4m + n) - (4m + 2n)

Left side: 12a + 2b + 56

Right side: -n

Therefore:

12a + 2b + 56 = -n

Let me call this Equation 6: 12a + 2b + 56 = -n

Similarly, let me look at equation 1: a + b + 6 = m + n. Let me call this Equation 1.

So, if I can express m and n in terms of a and b, maybe?

Alternatively, let me try to find another equation. Let me also consider equation 5: 5a + b + 19 = 2m.

So, if I can express m from equation 5: m = (5a + b + 19)/2.

Similarly, from equation 1: m + n = a + b + 6 => n = a + b + 6 - m.

So, substituting m from equation 5 into this:

n = a + b + 6 - (5a + b + 19)/2

Let me compute that:

Multiply numerator:

n = [2(a + b + 6) - 5a - b - 19]/2

Compute numerator:

2a + 2b + 12 - 5a - b - 19 = (-3a + b - 7)

Therefore, n = (-3a + b - 7)/2. Let me call this Equation 7.

Now, Equation 6 is 12a + 2b + 56 = -n. So, substituting n from Equation 7:

12a + 2b + 56 = - [ (-3a + b - 7)/2 ]

Multiply both sides by 2 to eliminate denominator:

2*(12a + 2b + 56) = 3a - b + 7

Left side: 24a + 4b + 112

So:

24a + 4b + 112 = 3a - b + 7

Bring all terms to left side:

24a + 4b + 112 - 3a + b - 7 = 0

Simplify:

21a + 5b + 105 = 0

So, 21a + 5b = -105. Let me call this Equation 8.

Now, let me see if I can find another equation involving a and b. Let me check equation 5: 5a + b + 19 = 2m. But we also have equation 1 and equation 7. Wait, maybe I can use equation 3?

Wait, equation 3 was 9a + 3b + 32 = 6m + 2n. Let me try to use that. Let me express m and n in terms of a and b.

From equation 5: m = (5a + b + 19)/2

From equation 7: n = (-3a + b - 7)/2

So, substitute m and n into equation 3:

Left side: 9a + 3b + 32

Right side: 6m + 2n = 6*(5a + b + 19)/2 + 2*(-3a + b - 7)/2

Simplify:

= 3*(5a + b + 19) + (-3a + b - 7)

= 15a + 3b + 57 - 3a + b - 7

= 12a + 4b + 50

Therefore, equation 3 becomes:

9a + 3b + 32 = 12a + 4b + 50

Bring all terms to left side:

9a + 3b + 32 - 12a - 4b - 50 = 0

Simplify:

-3a - b - 18 = 0 => -3a - b = 18 => 3a + b = -18. Let me call this Equation 9.

Now, we have Equation 8: 21a + 5b = -105 and Equation 9: 3a + b = -18.

So, we can solve these two equations. Let me solve Equation 9 for b: b = -18 - 3a. Then substitute into Equation 8:

21a + 5*(-18 - 3a) = -105

Compute:

21a - 90 - 15a = -105

Simplify:

6a - 90 = -105

Add 90 to both sides:

6a = -15 => a = -15/6 = -5/2.

So, a = -5/2. Then from Equation 9: b = -18 - 3*(-5/2) = -18 + 15/2 = (-36/2 + 15/2) = (-21)/2 = -10.5 or -21/2.

So, a = -5/2, b = -21/2.

Now, let me check if these values satisfy Equation 8:

21a + 5b = 21*(-5/2) + 5*(-21/2) = (-105/2) + (-105/2) = -210/2 = -105. Yes, that's correct.

So, now we have a and b. Now, let me find m and n.

From equation 5: m = (5a + b + 19)/2.

Plugging in a = -5/2 and b = -21/2:

5a = 5*(-5/2) = -25/2

b = -21/2

So, 5a + b + 19 = (-25/2 -21/2) + 19 = (-46/2) + 19 = -23 + 19 = -4

Therefore, m = -4 / 2 = -2.

So, m = -2.

From equation 7: n = (-3a + b - 7)/2.

Compute:

-3a = -3*(-5/2) = 15/2

b = -21/2

So, -3a + b -7 = 15/2 -21/2 -7 = (-6/2) -7 = -3 -7 = -10

Therefore, n = -10 / 2 = -5.

So, n = -5.

Therefore, R(x) = mx + n = -2x -5.

Let me check if this is consistent with the earlier equations.

For example, check equation 1: a + b + 6 = m + n.

a + b + 6 = (-5/2) + (-21/2) + 6 = (-26/2) + 6 = -13 + 6 = -7

m + n = -2 + (-5) = -7. Correct.

Similarly, check equation 2: 4a + 2b + 13 = 4m + 2n.

Compute left side: 4*(-5/2) + 2*(-21/2) + 13 = (-10) + (-21) + 13 = -18

Right side: 4*(-2) + 2*(-5) = -8 -10 = -18. Correct.

Equation 3: 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. Correct.

Equation 4: 16a + 4b + 69 = 4m + n.

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 is:

P(x) = x³ - (5/2)x² - (21/2)x + 5.

But maybe I should write it with fractions or convert to integers? Let me check if I can write it as:

Multiply through by 2 to eliminate denominators? But since the problem asks for P(5), maybe I can compute it directly.

Alternatively, maybe I can write P(x) as:

P(x) = x³ - (5/2)x² - (21/2)x + 5.

But let me check if there's another way to find P(5) without having to compute the entire polynomial? Maybe using the remainder information?

Wait, but since we have the polynomial, maybe it's easier to compute P(5) directly. Let me do that.

Compute P(5):

P(5) = 5³ - (5/2)(5²) - (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 = 15.

Wait, 125 - 115 is 10, plus 5 is 15. So P(5) = 15?

But let me check my calculation again to be sure.

Alternatively, maybe I made a mistake in the calculation. Let me recompute:

P(5) = 5³ - (5/2)(5²) - (21/2)(5) + 5

First term: 125

Second term: (5/2)(25) = 125/2 = 62.5

Third term: (21/2)(5) = 105/2 = 52.5

Fourth term: 5

So:

125 - 62.5 - 52.5 + 5

Compute 125 - 62.5 = 62.5

Then 62.5 - 52.5 = 10

Then 10 + 5 = 15. Yes, same result. So P(5) = 15.

But let me check if there's another way to get this result, maybe using the remainders?

Alternatively, since we know R(x) = -2x -5, and we know that P(x) = (x - 1)(x - 4)Q(x) + R(x). Also, since P(x) is cubic, Q(x) is linear. Let me write Q(x) as 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 of (x - 1)(x - 4)(px + q) is x² * px = px³. Since P(x) is monic, the coefficient of x³ is 1, so p must be 1. Therefore, Q(x) = x + q.

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

Let me expand this expression and compare with the previous expression for P(x).

First, expand (x - 1)(x - 4):

(x - 1)(x - 4) = x² - 5x + 4

Multiply by (x + q):

(x² - 5x + 4)(x + q) = x³ + qx² -5x² -5qx + 4x + 4q

Simplify:

x³ + (q - 5)x² + (-5q + 4)x + 4q

Then subtract 2x + 5:

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

Combine like terms:

x³ + (q - 5)x² + [(-5q + 4) - 2]x + (4q -5)

Simplify:

x³ + (q - 5)x² + (-5q + 2)x + (4q -5)

But we also have from earlier that P(x) = x³ - (5/2)x² - (21/2)x + 5. Therefore, equate coefficients:

For x³: 1 = 1, which is 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:

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

For constant term: 4q -5 = 4*(5/2) -5 = 10 -5 = 5. Correct. Perfect.

Therefore, q = 5/2. Therefore, P(x) can be written as:

(x - 1)(x - 4)(x + 5/2) - 2x -5.

But since we already found P(5) = 15, maybe we can check this expression for P(5):

Compute P(5) = (5 - 1)(5 - 4)(5 + 5/2) - 2*5 -5

Compute each part:

(5 - 1) = 4

(5 - 4) = 1

(5 + 5/2) = 15/2

Multiply them: 4 * 1 * 15/2 = 4*(15/2) = 30

Then subtract 2*5 +5 = 10 +5 = 15. Therefore, P(5) = 30 - 15 = 15. Correct. So that's another way to compute it.

Therefore, the answer is 15. But let me check once more with the original polynomial.

Original polynomial: P(x) = x³ - (5/2)x² - (21/2)x + 5.

Compute P(5):

5³ = 125

(5/2)x² at x=5: (5/2)*(25) = 125/2

(21/2)x at x=5: (21/2)*5 = 105/2

So:

P(5) = 125 - 125/2 - 105/2 + 5 = 125 - (125 + 105)/2 + 5 = 125 - 230/2 + 5 = 125 - 115 + 5 = 15. Correct.

Therefore, the answer is 15. However, let me check if there's another way to get this result without computing the entire polynomial, maybe using the given conditions?

Alternatively, since we know that P(x) = (x - 1)(x - 4)Q(x) + R(x), and also P(x) = (x - 2)(x - 3)S(x) + 2R(x). Since we know R(x) = -2x -5, then maybe we can use these expressions?

But since we already have the answer, maybe that's not necessary. However, just to ensure that there are no mistakes in the process, let me check if P(0) = 5. From the polynomial:

P(0) = 0³ - (5/2)(0)² - (21/2)(0) + 5 = 5. Correct. So that's good.

Another check: Let me compute P(1). From the polynomial:

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

Similarly, P(4) = 64 - (5/2)(16) - (21/2)(4) +5 = 64 - 40 - 42 +5 = 64 - 82 +5 = -13. Which should be R(4) = -2(4) -5 = -8 -5 = -13. Correct.

Similarly, P(2) = 8 - (5/2)(4) - (21/2)(2) +5 = 8 - 10 -21 +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 - 45/2 -63/2 +5 = 27 +5 - (45 +63)/2 = 32 - 108/2 = 32 -54 = -22. Which should be 2R(3) = 2*(-2*3 -5) = 2*(-6 -5) = 2*(-11) = -22. Correct.

All checks out. Therefore, P(5) = 15 is correct.

But just to make sure, let me think if there's another approach. For example, using the Chinese Remainder Theorem or something else?

Alternatively, since 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. Since P(x) is cubic, A(x) and B(x) are linear.

But since P(x) is monic, then A(x) must be monic? Let me see. If P(x) - R(x) is (x - 1)(x - 4)A(x), then the leading term of P(x) is x³, and the leading term of (x - 1)(x - 4)A(x) is x² * A(x)'s leading term. Since P(x) is monic, then A(x) must be monic. Similarly for B(x). Therefore, A(x) = x + c, B(x) = x + d.

But maybe this approach is similar to what I did before. However, since we already have the answer, maybe it's redundant.

Alternatively, since we know R(x) = -2x -5, then P(x) = (x - 1)(x - 4)Q(x) -2x -5. Also, since P(x) is monic cubic, Q(x) is linear, as before. So, if I can find Q(x), but we already did that.

Alternatively, since we have P(5), maybe we can use the fact that P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, P(5) - R(5) is divisible by (5 - 1)(5 - 4) = 4*1 = 4. Similarly, P(5) - 2R(5) is divisible by (5 - 2)(5 - 3) = 3*2 = 6. But since we already know P(5) = 15, let me check:

R(5) = -2*5 -5 = -15. Therefore, P(5) - R(5) = 15 - (-15) = 30, which is divisible by 4? 30 divided by 4 is 7.5, which is not integer. Wait, but that contradicts? Wait, but P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, P(5) - R(5) should be divisible by (5 - 1)(5 - 4) = 4. But 30 is not divisible by 4. Wait, that seems contradictory. What's wrong here?

Wait, hold on. Wait, if P(x) - R(x) is divisible by (x - 1)(x - 4), then P(x) - R(x) = (x - 1)(x - 4)Q(x). Therefore, for any x, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, in particular, when x = 5, P(5) - R(5) is divisible by (5 - 1)(5 - 4) = 4. But according to our calculation, P(5) - R(5) = 15 - (-15) = 30. 30 divided by 4 is 7.5, which is not integer. That seems contradictory. Wait, but this suggests that there is a mistake in my previous reasoning?

Wait, but earlier checks showed that P(x) is correct. So why is this discrepancy?

Wait, maybe my mistake is in assuming that P(x) - R(x) is divisible by (x - 1)(x - 4). Wait, the problem says that when P(x) is divided by (x - 1)(x - 4), the remainder is R(x). Therefore, P(x) = (x - 1)(x - 4)Q(x) + R(x). Therefore, P(x) - R(x) = (x - 1)(x - 4)Q(x), which is divisible by (x - 1)(x - 4). Therefore, for any x, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, in particular, when x = 5, P(5) - R(5) must be divisible by (5 - 1)(5 - 4) = 4. However, according to our calculation, P(5) - R(5) = 15 - (-15) = 30, which is not divisible by 4. Therefore, there must be a mistake here. Wait, but earlier checks showed that P(x) is correct. So where is the error?

Wait, let me check R(x). Earlier, we found R(x) = -2x -5. Let me check if that is correct. From the equations, we had:

P(1) = -7, which is R(1) = -2(1) -5 = -7. Correct.

Similarly, P(4) = -13, which is R(4) = -2(4) -5 = -13. Correct.

Therefore, R(x) is correct. Then P(5) - R(5) = 15 - (-15) = 30. But 30 divided by 4 is 7.5. But according to the division algorithm, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, P(5) - R(5) must be divisible by (5 - 1)(5 - 4) = 4. But 30 is not divisible by 4. Therefore, there is a contradiction here. Which means that there is a mistake in my previous calculations. Wait, but all the checks for P(1), P(2), P(3), P(4) were correct. How is this possible?

Wait, let me check P(5) again. If P(x) = x³ - (5/2)x² - (21/2)x + 5, then P(5) is:

5³ = 125

(5/2)x² at x=5: (5/2)(25) = 125/2 = 62.5

(21/2)x at x=5: (21/2)(5) = 105/2 = 52.5

So P(5) = 125 - 62.5 - 52.5 + 5 = 125 - 115 + 5 = 15. Correct.

R(5) = -2*5 -5 = -15. Therefore, P(5) - R(5) = 15 - (-15) = 30. But 30 divided by 4 is 7.5. But according to the division algorithm, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, P(5) - R(5) should be divisible by (5 - 1)(5 - 4) = 4. But 30 is not divisible by 4. Therefore, there is a contradiction. Which means that there is a mistake in my previous steps. But where?

Wait, let me check the equations again. Let me check the equations for P(1), P(2), P(3), P(4). Earlier, I had:

From the polynomial P(x) = x³ - (5/2)x² - (21/2)x + 5.

Compute P(1):

1 - (5/2) - (21/2) + 5 = 1 - 26/2 + 5 = 1 -13 +5 = -7. Correct.

P(2):

8 - (5/2)(4) - (21/2)(2) +5 = 8 - 10 -21 +5 = (8 +5) - (10 +21) = 13 - 31 = -18. Which is 2R(2) = 2*(-2*2 -5) = 2*(-9) = -18. Correct.

P(3):

27 - (5/2)(9) - (21/2)(3) +5 = 27 - 45/2 -63/2 +5 = 27 +5 - (45 +63)/2 = 32 - 108/2 = 32 -54 = -22. Which is 2R(3) = 2*(-2*3 -5) = 2*(-11) = -22. Correct.

P(4):

64 - (5/2)(16) - (21/2)(4) +5 = 64 - 40 -42 +5 = 64 - 82 +5 = -13. Which is R(4) = -2*4 -5 = -13. Correct.

So all these are correct. Then why is P(5) - R(5) not divisible by 4?

Wait, maybe my mistake is in the assumption that P(x) - R(x) is divisible by (x - 1)(x - 4). Wait, the division algorithm says that when you divide P(x) by (x - 1)(x - 4), you get quotient Q(x) and remainder R(x). Therefore, P(x) = (x - 1)(x - 4)Q(x) + R(x). Therefore, P(x) - R(x) = (x - 1)(x - 4)Q(x). Therefore, for any x, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, in particular, when x = 5, P(5) - R(5) must be divisible by (5 - 1)(5 - 4) = 4. But according to our calculation, P(5) - R(5) = 30, which is not divisible by 4. Therefore, there is a contradiction. Therefore, my previous solution must have an error.

Wait, but how? All the checks for P(1), P(2), P(3), P(4) are correct. Let me check the value of Q(x). Earlier, I said that Q(x) = x + q, with q = 5/2. Therefore, P(x) = (x - 1)(x - 4)(x + 5/2) - 2x -5. Let me compute P(5) using this expression:

(x - 1)(x - 4)(x + 5/2) - 2x -5 at x = 5:

(5 - 1)(5 - 4)(5 + 5/2) - 2*5 -5 = 4*1*(15/2) - 10 -5 = 30 - 15 = 15. Correct. So P(5) = 15. Then R(5) = -2*5 -5 = -15. Therefore, P(5) - R(5) = 15 - (-15) = 30. But according to the division algorithm, this should be divisible by (5 - 1)(5 - 4) = 4. But 30 is not divisible by 4. What is happening here?

Wait, maybe my mistake is in the definition of R(x). Wait, the problem says that when P(x) is divided by (x - 1)(x - 4), the remainder is R(x). But R(x) is a linear polynomial. However, when we divide P(x) by (x - 1)(x - 4), the remainder is R(x), which is of degree less than 2, so degree 1 or 0. But in our case, R(x) is linear. However, when we compute P(x) - R(x), it is divisible by (x - 1)(x - 4). Therefore, P(x) - R(x) = (x - 1)(x - 4)Q(x). Therefore, for any x, P(x) - R(x) is divisible by (x - 1)(x - 4). Therefore, in particular, when x = 5, P(5) - R(5) must be divisible by (5 - 1)(5 - 4) = 4. But 30 is not divisible by 4. Therefore, there is a contradiction. Therefore, my previous solution is wrong. But how?

Wait, but all the checks for P(1), P(2), P(3), P(4) are correct. Let me check if P(x) - R(x) is divisible by (x - 1)(x - 4). Let me compute P(x) - R(x):

P(x) - R(x) = x³ - (5/2)x² - (21/2)x + 5 - (-2x -5) = x³ - (5/2)x² - (21/2)x + 5 + 2x + 5 = x³ - (5/2)x² - (17/2)x + 10.

Now, let me check if this polynomial is divisible by (x - 1)(x - 4) = x² - 5x + 4.

Let me perform polynomial division or factor it.

Let me try to factor x³ - (5/2)x² - (17/2)x + 10.

Alternatively, since we know that (x - 1)(x - 4) is a factor, then it should be divisible by x² - 5x + 4. Let me perform the division.

Divide x³ - (5/2)x² - (17/2)x + 10 by x² - 5x + 4.

Let me write it as:

Dividend: x³ - (5/2)x² - (17/2)x + 10

Divisor: x² - 5x + 4

First term: x³ / x² = x. Multiply divisor by x: x³ - 5x² + 4x.

Subtract this from dividend:

(x³ - (5/2)x² - (17/2)x + 10) - (x³ - 5x² + 4x) = [0x³] + [(-5/2 + 5)x²] + [(-17/2 - 4)x] + 10

Compute coefficients:

-5/2 + 5 = -5/2 + 10/2 = 5/2

-17/2 - 4 = -17/2 - 8/2 = -25/2

So, the remainder is (5/2)x² - (25/2)x + 10.

Now, divide this remainder by the divisor x² - 5x + 4.

Next term: (5/2)x² / x² = 5/2. Multiply divisor by 5/2: (5/2)x² - (25/2)x + 10.

Subtract this from the remainder:

(5/2)x² - (25/2)x + 10 - [(5/2)x² - (25/2)x + 10] = 0. Perfect, no remainder.

Therefore, P(x) - R(x) = (x - 1)(x - 4)(x + 5/2). Therefore, it is divisible by (x - 1)(x - 4). Therefore, there is no contradiction. So why is P(5) - R(5) = 30 not divisible by 4?

Wait, because P(x) - R(x) is divisible by (x - 1)(x - 4), which means that for any x, P(x) - R(x) is divisible by (x - 1)(x - 4). But when x = 5, (x - 1)(x - 4) = 4*1 = 4. Therefore, P(5) - R(5) must be divisible by 4. But 30 is not divisible by 4. Wait, but according to the division, P(x) - R(x) = (x - 1)(x - 4)(x + 5/2). Therefore, P(5) - R(5) = (5 - 1)(5 - 4)(5 + 5/2) = 4*1*(15/2) = 30. Which is equal to 30. But 30 is divisible by 4? No, 30 divided by 4 is 7.5. Wait, but according to the polynomial division, P(x) - R(x) is divisible by (x - 1)(x - 4), so P(x) - R(x) = (x - 1)(x - 4)Q(x). Therefore, when x = 5, P(5) - R(5) = (5 - 1)(5 - 4)Q(5) = 4*1*Q(5). Therefore, P(5) - R(5) = 4*Q(5). Therefore, it must be divisible by 4. But according to our calculation, P(5) - R(5) = 30, which would mean that Q(5) = 30 / 4 = 7.5. But Q(x) is a linear polynomial, which is x + 5/2. Therefore, Q(5) = 5 + 5/2 = 15/2 = 7.5. Which is correct. Therefore, 4 * 7.5 = 30. So, 30 is divisible by 4? No, but 30 is equal to 4 * 7.5. But 7.5 is not an integer. However, the division algorithm doesn't require Q(x) to have integer coefficients. Q(x) is a polynomial with real coefficients, not necessarily integer. Therefore, there is no contradiction. My mistake was assuming that Q(x) must have integer coefficients, but it doesn't. Therefore, everything is correct. So P(5) = 15 is correct.

Therefore, the answer is 15. So I think my initial solution is correct, and the confusion was due to my incorrect assumption about Q(x) having integer coefficients. Therefore, the final answer is 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: General Form of the Polynomial

Since $ P(x) $ is a monic cubic polynomial, we can write it as:

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

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

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

Thus, the polynomial becomes:

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

---

### Step 2: Use the Remainder Theorem

Let $ R(x) = mx + n $, a linear polynomial (since the remainder when dividing by a quadratic is of degree less than 2).

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) $:

- $ 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 $

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

1. $ a + b + 6 = m + n $
2. $ 4a + 2b + 13 = 4m + 2n $
3. $ 9a + 3b + 32 = 6m + 2n $
4. $ 16a + 4b + 69 = 4m + n $

---

### Step 3: Solve the System of Equations

From equations (1) and (2), subtract (1) from (2):

$$
(4a + 2b + 13) - (a + b + 6) = (4m + 2n) - (m + n)
\Rightarrow 3a + b + 7 = 3m + n \tag{5}
$$

From equations (2) and (4), subtract (2) from (4):

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

From equation (5), solve for $ n $:

$$
n = 3a + b + 7 - 3m \tag{7}
$$

Substitute (7) into (6):

$$
12a + 2b + 56 = - (3a + b + 7 - 3m)
\Rightarrow 12a + 2b + 56 = -3a - b - 7 + 3m
\Rightarrow 15a + 3b + 63 = 3m \tag{8}
$$

Now, from equation (1):

$$
a + b + 6 = m + n \Rightarrow n = a + b + 6 - m \tag{9}
$$

Substitute (9) into (6):

$$
12a + 2b + 56 = - (a + b + 6 - m)
\Rightarrow 12a + 2b + 56 = -a - b - 6 + m
\Rightarrow 13a + 3b + 62 = m \tag{10}
$$

Now, from (8) and (10), equate expressions for $ m $:

$$
15a + 3b + 63 = 13a + 3b + 62
\Rightarrow 2a = -1 \Rightarrow a = -\frac{1}{2}
$$

Substitute $ a = -\frac{1}{2} $ into (10):

$$
13(-\frac{1}{2}) + 3b + 62 = m
\Rightarrow -\frac{13}{2} + 3b + 62 = m
\Rightarrow 3b + \frac{101}{2} = m
$$

Now substitute into (1):

$$
a + b + 6 = m + n
\Rightarrow -\frac{1}{2} + b + 6 = m + n
\Rightarrow b + \frac{11}{2} = m + n
$$

From (9): $ n = a + b + 6 - m = -\frac{1}{2} + b + 6 - m = b + \frac{11}{2} - m $

So:

$$
b + \frac{11}{2} = m + (b + \frac{11}{2} - m) \Rightarrow \text{True}
$$

Now, from (10): $ m = 3b + \frac{101}{2} $

Substitute into (1):

$$
a + b + 6 = m + n = m + (b + \frac{11}{2} - m) = b + \frac{11}{2}
\Rightarrow -\frac{1}{2} + b + 6 = b + \frac{11}{2}
\Rightarrow \frac{11}{2} = \frac{11}{2}
$$

So the system is consistent.

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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,
\[P(5) = (5 + p)(4)(1) - 10 + d = 10 + 4p + d = \boxed{15}.\]