Explanation:
Okay, let's tackle this problem step by step. So, we have a monic polynomial P(x) of degree 3. That means the leading coefficient is 1. 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, and we need to find P(5).

First, let me recall the Remainder Factor Theorem and polynomial division concepts. When a polynomial P(x) is divided by a quadratic polynomial like (x - a)(x - b), the remainder will be a linear polynomial (degree less than 2). So R(x) must be of the form Ax + B, where A and B are constants. Similarly, 2R(x) would then be 2Ax + 2B.

So, since P(x) divided by (x - 1)(x - 4) leaves remainder R(x) = Ax + B, we can write:

P(x) = (x - 1)(x - 4)Q(x) + Ax + B

for some polynomial Q(x). Similarly, when divided by (x - 2)(x - 3), the remainder is 2R(x) = 2Ax + 2B, so:

P(x) = (x - 2)(x - 3)S(x) + 2Ax + 2B

for some polynomial S(x). Since P(x) is a cubic polynomial, and (x - 1)(x - 4) is quadratic, Q(x) must be linear. Similarly, S(x) is also linear.

But maybe there's a better way to approach this without dealing with Q(x) and S(x). Let me think. Since we know that when we divide by (x - 1)(x - 4), the remainder is R(x), then evaluating P(x) at x = 1 and x = 4 should give us the remainder evaluated at those points. Similarly for the other division.

Yes, that seems right. Because if we have P(x) = (x - 1)(x - 4)Q(x) + R(x), then when x = 1, the term (x - 1)(x - 4)Q(x) becomes zero, so P(1) = R(1). Similarly, P(4) = R(4). Similarly, for the other division, P(2) = 2R(2) and P(3) = 2R(3).

Therefore, we can write:

P(1) = R(1)
P(4) = R(4)
P(2) = 2R(2)
P(3) = 2R(3)

But R(x) is Ax + B, so:

R(1) = A(1) + B = A + B
R(2) = 2A + B
R(3) = 3A + B
R(4) = 4A + B

Therefore:

P(1) = A + B
P(4) = 4A + B
P(2) = 2(2A + B) = 4A + 2B
P(3) = 2(3A + B) = 6A + 2B

Additionally, since P(x) is a cubic polynomial, we can express it in terms of its coefficients. Since it's monic, the leading term is x^3. Let me denote P(x) as:

P(x) = x^3 + cx^2 + dx + e

But maybe it's better to use the fact that we can write P(x) in terms of the remainders. Alternatively, since we have four equations (from P(1), P(2), P(3), P(4)), and we also know P(0) = 5. Wait, so P(0) is e, right? If P(x) is x^3 + cx^2 + dx + e, then P(0) = e = 5. So that's another equation.

Therefore, if we can find P(1), P(2), P(3), P(4), then we can set up equations. However, since we don't know the values of P(1), P(2), P(3), P(4) yet, but they are related to A and B through those equations. Let me see.

But also, since we have P(x) as a cubic polynomial, maybe we can use the fact that P(x) - R(x) is divisible by (x - 1)(x - 4), and P(x) - 2R(x) is divisible by (x - 2)(x - 3). Therefore, (x - 1)(x - 4) divides P(x) - R(x), and (x - 2)(x - 3) divides P(x) - 2R(x).

But since P(x) is a cubic, and R(x) is linear, then P(x) - R(x) is a cubic polynomial. But (x - 1)(x - 4) is quadratic, so P(x) - R(x) is divisible by that quadratic, meaning that P(x) - R(x) has roots at x = 1 and x = 4. Similarly, P(x) - 2R(x) has roots at x = 2 and x = 3.

But maybe this isn't helping me directly. Let me think again. Let's try to write equations using the information we have.

We have P(1) = A + B, P(4) = 4A + B, P(2) = 4A + 2B, P(3) = 6A + 2B. Also, P(0) = 5.

But since P(x) is a cubic polynomial, we can write down its general form. Let me try that. Let me assume:

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

Given that P(0) = c = 5. Therefore, c = 5.

Therefore, P(x) = x^3 + ax^2 + bx + 5.

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

Compute P(1) = 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 earlier, we have:

P(1) = A + B

P(4) = 4A + B

P(2) = 4A + 2B

P(3) = 6A + 2B

So let's write equations:

From P(1) = A + B = a + b + 6    (Equation 1)

From P(4) = 4A + B = 16a + 4b + 69   (Equation 2)

From P(2) = 4A + 2B = 4a + 2b + 13    (Equation 3)

From P(3) = 6A + 2B = 9a + 3b + 32    (Equation 4)

So now we have four equations with variables A, B, a, b. However, we can solve for A and B first using Equations 1, 2, 3, 4, and then relate them to a and b.

Alternatively, since Equations 1, 2, 3, 4 might be redundant. Let me check.

First, let me note that Equations 1 and 3 can be used to solve for A and B in terms of a and b. Similarly, Equations 1 and 2, or Equations 3 and 4. Let me try with Equations 1 and 3 first.

From Equation 1: A + B = a + b + 6

From Equation 3: 4A + 2B = 4a + 2b + 13

Let me write Equation 3 as 2*(2A + B) = 2*(2a + b) + 13? Wait, maybe better to solve the equations:

Let me write:

Equation 1: A + B = a + b + 6

Equation 3: 4A + 2B = 4a + 2b + 13

Let me denote Equation 1 multiplied by 2: 2A + 2B = 2a + 2b + 12

Subtract this from Equation 3:

(4A + 2B) - (2A + 2B) = (4a + 2b + 13) - (2a + 2b + 12)

Which gives:

2A = 2a + 1

Therefore, A = a + 0.5

Then, substitute back into Equation 1: A + B = a + b + 6 => (a + 0.5) + B = a + b + 6 => B = b + 6 - 0.5 => B = b + 5.5

Therefore, A = a + 1/2, B = b + 11/2.

Now, let's check if these hold with Equation 4. Equation 4 is:

6A + 2B = 9a + 3b + 32

Substitute A and B:

6*(a + 1/2) + 2*(b + 11/2) = 9a + 3b + 32

Compute left side:

6a + 3 + 2b + 11 = 6a + 2b + 14

Set equal to right side:

6a + 2b + 14 = 9a + 3b + 32

Bring all terms to left:

6a + 2b +14 -9a -3b -32 = 0 => -3a - b -18 = 0 => 3a + b = -18 (Equation 5)

Similarly, check Equation 2: 4A + B = 16a + 4b + 69

Substitute A and B:

4*(a + 1/2) + (b + 11/2) = 4a + 2 + b + 5.5 = 4a + b + 7.5

Set equal to 16a + 4b + 69:

4a + b + 7.5 = 16a + 4b + 69

Bring all terms to left:

4a + b + 7.5 -16a -4b -69 = 0 => -12a -3b -61.5 = 0 => Multiply both sides by 2 to eliminate decimal:

-24a -6b -123 = 0 => 24a + 6b = -123 => Divide both sides by 3: 8a + 2b = -41 (Equation 6)

So now we have two equations:

Equation 5: 3a + b = -18

Equation 6: 8a + 2b = -41

Let me solve these two equations. Let me use Equation 5 to express b = -18 - 3a. Then substitute into Equation 6.

Equation 6: 8a + 2*(-18 - 3a) = -41

Compute:

8a - 36 - 6a = -41 => 2a - 36 = -41 => 2a = -41 + 36 => 2a = -5 => a = -5/2

Then, from Equation 5: 3*(-5/2) + b = -18 => -15/2 + b = -18 => b = -18 + 15/2 = (-36/2 + 15/2) = -21/2

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

Therefore, we have found a and b. Now, we can find A and B from earlier:

A = a + 1/2 = (-5/2) + 1/2 = (-4/2) = -2

B = b + 11/2 = (-21/2) + 11/2 = (-10/2) = -5

So, R(x) = Ax + B = -2x -5

Therefore, R(x) = -2x -5.

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

For example, P(1) = A + B = -2*1 -5 = -7. Let me compute P(1) using the polynomial:

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

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

Similarly, check P(2):

P(2) = 8 - (5/2)(4) - (21/2)(2) + 5 = 8 - 10 - 21 + 5 = (8 + 5) - (10 + 21) = 13 - 31 = -18.

But according to earlier, P(2) should be 4A + 2B = 4*(-2) + 2*(-5) = -8 -10 = -18. Correct.

Similarly, check P(4):

P(4) = 64 - (5/2)(16) - (21/2)(4) + 5 = 64 - 40 - 42 + 5 = (64 + 5) - (40 + 42) = 69 - 82 = -13.

According to Equation 2: 4A + B = 4*(-2) + (-5) = -8 -5 = -13. Correct.

Also check P(3):

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

According to Equation 4: 6A + 2B = 6*(-2) + 2*(-5) = -12 -10 = -22. Correct.

So, all equations check out. Therefore, our found values for a, b, A, B are correct.

So now, the polynomial P(x) is:

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

But perhaps we can write this with fractions combined:

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

Alternatively, to make it easier for computation, maybe multiply through by 2 to eliminate denominators:

But since we need to find P(5), maybe we can compute it directly.

Alternatively, maybe we can write P(x) in terms of its roots or something else? Not sure. Let me compute P(5) directly.

Compute P(5):

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

Compute each term:

5^3 = 125

(5/2)(5)^2 = (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) + 5 = 10 + 5 = 15.

Wait, so is P(5) equal to 15? Let me check the calculation again step by step to be sure.

First term: 5^3 = 125

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

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

Fourth term: +5

Therefore, adding all together:

125 - 125/2 - 105/2 + 5

Convert 125 and 5 to halves:

125 = 250/2, and 5 = 10/2

Therefore:

250/2 - 125/2 - 105/2 + 10/2 = (250 - 125 - 105 + 10)/2 = (250 - 230 + 10)/2 = (20)/2 = 10.

Wait, hold on. Wait, 250 - 125 is 125, then 125 - 105 is 20, then 20 + 10 is 30? Wait, no:

Wait, let me recast:

Original terms:

125 - 125/2 - 105/2 + 5

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

= 130 - (230/2)

= 130 - 115 = 15

Yes, so earlier step where I converted to halves:

Original:

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

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

= 130 - (230/2)

= 130 - 115 = 15.

Yes, so my first calculation was correct. So P(5) = 15.

But let me double-check by plugging into the polynomial:

P(x) = x³ - (5/2)x² - (21/2)x + 5

So P(5) = 125 - (5/2)(25) - (21/2)(5) + 5 = 125 - 125/2 - 105/2 + 5.

Compute 125 - 125/2 is 125*(1 - 1/2) = 125*(1/2) = 62.5. Then subtract 105/2 which is 52.5. So 62.5 - 52.5 = 10. Then add 5: 10 + 5 = 15. Yep, same result. So P(5) is 15.

Therefore, the answer should be 15.

But wait, let me check again if there's another way to approach this problem, maybe using interpolation or something else, just to confirm.

Alternatively, since we know that R(x) = -2x -5, then when P(x) is divided by (x -1)(x -4), remainder is -2x -5. Also, since the remainder when divided by (x -1)(x -4) is R(x), then P(x) is equal to (x -1)(x -4)Q(x) + R(x). Similarly for the other division. So, since we also have P(0) = 5, which we used to find c = 5, which helped us find the coefficients a and b.

Alternatively, since we have R(x) = -2x -5, and we also have that P(x) - R(x) is divisible by (x -1)(x -4). Therefore, P(x) - (-2x -5) = P(x) + 2x +5 is divisible by (x -1)(x -4). Similarly, P(x) - 2R(x) = P(x) - 2*(-2x -5) = P(x) +4x +10 is divisible by (x -2)(x -3).

Therefore, we can write:

P(x) + 2x +5 = (x -1)(x -4)(some linear polynomial)

Similarly,

P(x) +4x +10 = (x -2)(x -3)(some linear polynomial)

Since P(x) is cubic, then those "some linear polynomials" would be linear. Let me denote:

Let me call the first one: P(x) + 2x +5 = (x -1)(x -4)(mx + n)

Similarly, P(x) +4x +10 = (x -2)(x -3)(px + q)

But since P(x) is same in both, we can set up equations. However, since we already found P(x), maybe this is redundant. But if we didn't know P(x), maybe this could be another way.

Alternatively, let me try plugging x = 1 into P(x) + 2x +5. Since (x -1) divides P(x) +2x +5, then at x = 1, P(1) + 2*1 +5 = 0. Wait, no, since P(x) + 2x +5 is divisible by (x -1), so P(1) + 2*1 +5 should be zero? Wait, no. If P(x) + 2x +5 is divisible by (x -1), then when x = 1, P(1) + 2*1 +5 = 0.

But we know that P(1) = R(1) = -2*1 -5 = -7. Then:

-7 + 2 + 5 = -7 +7 = 0. Correct. Similarly, at x = 4, P(4) + 2*4 +5 = 0. P(4) = R(4) = -8 -5 = -13. Then -13 + 8 +5 = 0. Correct.

Similarly, for the other equation, P(x) +4x +10 is divisible by (x -2)(x -3). So at x = 2, P(2) +4*2 +10 = 0. P(2) = -18. Then -18 + 8 +10 = 0. Correct. Similarly at x = 3, P(3) +4*3 +10 = -22 + 12 +10 = 0. Correct.

So that checks out. Therefore, this method also confirms our previous calculations.

Alternatively, if I didn't have P(x), maybe I could set up equations for the coefficients. But since we already have P(x), and verified all the conditions, and found P(5) = 15, which seems correct.

Wait, but let me check once more if there is a mistake in calculating the coefficients.

We had:

From Equation 5: 3a + b = -18

From Equation 6: 8a + 2b = -41

We solved it by substituting b = -18 - 3a into Equation 6:

8a + 2*(-18 -3a) = 8a -36 -6a = 2a -36 = -41

So 2a = -5 => a = -5/2. Then b = -18 - 3*(-5/2) = -18 + 15/2 = -36/2 + 15/2 = -21/2. Correct.

Therefore, the coefficients are correct. Then the polynomial is:

x^3 - (5/2)x^2 - (21/2)x + 5.

So, if I want to write this polynomial with integer coefficients, I can multiply by 2:

2P(x) = 2x^3 -5x^2 -21x + 10

But maybe not necessary here.

Alternatively, maybe I can find P(5) using another approach. For example, since we know that P(x) - R(x) is divisible by (x -1)(x -4), so P(x) - R(x) = (x -1)(x -4)(something). Similarly, P(x) - 2R(x) = (x -2)(x -3)(something else).

But maybe we can write that P(x) - R(x) = (x -1)(x -4)(kx + m). Similarly, P(x) - 2R(x) = (x -2)(x -3)(lx + n). Since P(x) is cubic, then kx + m and lx + n are linear polynomials.

Alternatively, since we already have R(x) = -2x -5, then:

P(x) = (x -1)(x -4)(kx + m) -2x -5

But also, since P(x) is monic, the leading term of P(x) is x^3. Let me compute the leading term of the right-hand side:

(x -1)(x -4)(kx + m) = (x^2 -5x +4)(kx + m) = kx^3 + (m -5k)x^2 + (4k -5m)x + 4m

Therefore, when we subtract 2x +5, the leading term is kx^3. Since P(x) is monic, so k = 1. So:

P(x) = (x^2 -5x +4)(x + m) -2x -5

Let me expand this:

First multiply (x^2 -5x +4)(x + m):

= x^3 + m x^2 -5x^2 -5m x +4x +4m

= x^3 + (m -5)x^2 + (-5m +4)x + 4m

Then subtract 2x +5:

= x^3 + (m -5)x^2 + (-5m +4)x + 4m -2x -5

= x^3 + (m -5)x^2 + (-5m +4 -2)x + (4m -5)

= x^3 + (m -5)x^2 + (-5m + 2)x + (4m -5)

But we also have P(x) = x^3 - (5/2)x^2 - (21/2)x + 5. Therefore, equate coefficients:

For x^3: 1 = 1. Okay.

For x^2: m -5 = -5/2 => m = -5/2 +5 = 5/2.

For x term: -5m + 2 = -21/2. Let's check with m = 5/2:

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

For constant term: 4m -5 = 5. Let's check:

4*(5/2) -5 = 10 -5 = 5. Correct.

Therefore, m = 5/2. So that's consistent. Therefore, this approach also confirms that. Therefore, P(x) is indeed as we found earlier.

Therefore, P(5) = 15 is correct.

Alternatively, maybe there's a way to use the Chinese Remainder Theorem or something? But since we have two different remainders for two different divisors, but the problem is already solved through the above methods.

Alternatively, maybe use the fact that P(x) - R(x) has roots at 1 and 4, so P(1) = R(1) and P(4) = R(4), which we used.

But regardless, all checks confirm that P(5) = 15. Therefore, I think that's the correct answer.

**Final Answer**
\boxed{15}

We are given a **monic cubic polynomial** $ P(x) $ such that:

- 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) $.
- We are also told that $ P(0) = 5 $.
- We are to find $ P(5) $.

---

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

Let $ P(x) $ be a monic cubic polynomial:
$$
P(x) = x^3 + ax^2 + bx + c
$$

From the problem, we know:
$$
P(0) = c = 5
$$

So we can write:
$$
P(x) = x^3 + ax^2 + bx + 5
$$

---

### Step 2: Use the Remainder Theorem

From the Remainder Theorem, if a polynomial $ P(x) $ is divided by a factor $ (x - a) $, the remainder is simply $ P(a) $.

Since:
- $ P(x) $ mod $ (x - 1)(x - 4) $ gives remainder $ R(x) = Ax + B $,
- $ P(x) $ mod $ (x - 2)(x - 3) $ gives remainder $ 2R(x) = 2Ax + 2B $,

We can deduce:
$$
P(1) = R(1) = A + B \\
P(2) = 2R(2) = 2(2A + B) = 4A + 2B \\
P(3) = 2R(3) = 2(3A + B) = 6A + 2B \\
P(4) = R(4) = 4A + B
$$

So we have the following system of equations:

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

But we can also solve this using the fact that $ P(x) - R(x) $ is divisible by $ (x - 1)(x - 4) $, and $ P(x) - 2R(x) $ is divisible by $ (x - 2)(x - 3) $. That is:

- $ P(x) - R(x) = (x - 1)(x - 4)(mx + n) $
- $ P(x) - 2R(x) = (x - 2)(x - 3)(px + q) $

However, a more direct and simpler path is to solve the above system of four equations for $ A, B, a, b $, using substitution and elimination.

---

### Step 3: Solve the System of Equations

From equation (1):
$$
A + B = a + b + 6 \quad \text{(Equation 1)}
$$

From equation (3):
$$
4A + 2B = 4a + 2b + 13 \quad \text{(Equation 3)}
$$

Multiply Equation 1 by 2:
$$
2A + 2B = 2a + 2b + 12
$$

Subtract from Equation 3:
$$
(4A + 2B) - (2A + 2B) = (4a + 2b + 13) - (2a + 2b + 12)
$$
$$
2A = 2a + 1 \Rightarrow A = a + \frac{1}{2}
$$

Substitute into (1):
$$
A + B = a + b + 6 \Rightarrow a + \frac{1}{2} + B = a + b + 6 \Rightarrow B = b + 5.5
$$

Now use equation (4):
$$
6A + 2B = 9a + 3b + 32
$$

Substitute $ A = a + \frac{1}{2} $, $ B = b + \frac{11}{2} $:
$$
6(a + 0.5) + 2(b + 5.5) = 9a + 3b + 32
$$
$$
6a + 3 + 2b + 11 = 9a + 3b + 32
\Rightarrow 6a + 2b + 14 = 9a + 3b + 32
$$
$$
-3a - b = -18 \Rightarrow 3a + b = -18 \quad \text{(Equation 5)}
$$

Now consider equation (2):
$$
4A + B = 16a + 4b + 69
$$
Substitute $ A = a + \frac{1}{2} $, $ B = b + \frac{11}{2} $:
$$
4(a + 0.5) + b + \frac{11}{2} = 16a + 4b + 69
\Rightarrow 4a + 2 + b + 5.5 = 16a + 4b + 69
\Rightarrow 4a + b + 7.5 = 16a + 4b + 69
\Rightarrow -12a -3b = -61.5
\Rightarrow 8a + 2b = -41 \quad \text{(Equation 6)}
$$

Now solve the system:
$$
\begin{cases}
3a + b = -18 \\
8a + 2b = -41
\end{cases}
$$

From first equation:
$$
b = -18 - 3a
$$

Substitute into second:
$$
8a + 2(-18 - 3a) = -41 \Rightarrow 8a - 36 - 6a = -41 \Rightarrow 2a = -5 \Rightarrow a = -\frac{5}{2}
$$
$$
b = -18 - 3(-\frac{5}{2}) = -18 + \frac{15}{2} = -\frac{21}{2}
$$

So the polynomial becomes:
$$
P(x) = x^3 - \frac{5}{2}x^2 - \frac{21}{2}x + 5
$$

---

### Step 4: Compute $ P(5) $

$$
P(5) = 5^3 - \frac{5}{2}(5)^2 - \frac{21}{2}(5) + 5
= 125 - \frac{125}{2} - \frac{105}{2} + 5
= 125 - \frac{230}{2} + 5
= 125 - 115 + 5 = \boxed{15}
$$

---

### Final Answer

$$
\boxed{15}
$$

[Assistant]
Explanation:
Okay, let's tackle this problem step by step. So, we have a monic polynomial P(x) of degree 3. That means the leading coefficient is 1. 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, and we need to find P(5).

First, let me recall the Remainder Factor Theorem and polynomial division concepts. When a polynomial P(x) is divided by a quadratic polynomial like (x - a)(x - b), the remainder will be a linear polynomial (degree less than 2). So R(x) must be of the form Ax + B, where A and B are constants. Similarly, 2R(x) would then be 2Ax + 2B.

So, since P(x) divided by (x - 1)(x - 4) leaves remainder R(x) = Ax + B, we can write:

P(x) = (x - 1)(x - 4)Q(x) + Ax + B

for some polynomial Q(x). Similarly, when divided by (x - 2)(x - 3), the remainder is 2R(x) = 2Ax + 2B, so:

P(x) = (x - 2)(x - 3)S(x) + 2Ax + 2B

for some polynomial S(x). Since P(x) is a cubic polynomial, and (x - 1)(x - 4) is quadratic, Q(x) must be linear. Similarly, S(x) is also linear.

But maybe there's a better way to approach this without dealing with Q(x) and S(x). Let me think. Since we know that when we divide by (x - 1)(x - 4), the remainder is R(x), then evaluating P(x) at x = 1 and x = 4 should give us the remainder evaluated at those points. Similarly for the other division.

Yes, that seems right. Because if we have P(x) = (x - 1)(x - 4)Q(x) + R(x), then when x = 1, the term (x - 1)(x - 4)Q(x) becomes zero, so P(1) = R(1). Similarly, P(4) = R(4). Similarly, for the other division, P(2) = 2R(2) and P(3) = 2R(3).

Therefore, we can write:

P(1) = R(1)
P(4) = R(4)
P(2) = 2R(2)
P(3) = 2R(3)

But R(x) is Ax + B, so:

R(1) = A(1) + B = A + B
R(2) = 2A + B
R(3) = 3A + B
R(4) = 4A + B

Therefore:

P(1) = A + B
P(4) = 4A + B
P(2) = 2(2A + B) = 4A + 2B
P(3) = 2(3A + B) = 6A + 2B

Additionally, since P(x) is a cubic polynomial, we can express it in terms of its coefficients. Since it's monic, the leading term is x^3. Let me denote P(x) as:

P(x) = x^3 + cx^2 + dx + e

But maybe it's better to use the fact that we can write P(x) in terms of the remainders. Alternatively, since we have four equations (from P(1), P(2), P(3), P(4)), and we also know P(0) = 5. Wait, so P(0) is e, right? If P(x) is x^3 + cx^2 + dx + e, then P(0) = e = 5. So that's another equation.

Therefore, if we can find P(1), P(2), P(3), P(4), then we can set up equations. However, since we don't know the values of P(1), P(2), P(3), P(4) yet, but they are related to A and B through those equations. Let me see.

But also, since we have P(x) as a cubic polynomial, maybe we can use the fact that P(x) - R(x) is divisible by (x - 1)(x - 4), and P(x) - 2R(x) is divisible by (x - 2)(x - 3). Therefore, (x - 1)(x - 4) divides P(x) - R(x), and (x - 2)(x - 3) divides P(x) - 2R(x).

But since P(x) is a cubic, and R(x) is linear, then P(x) - R(x) is a cubic polynomial. But (x - 1)(x - 4) is quadratic, so P(x) - R(x) is divisible by that quadratic, meaning that P(x) - R(x) has roots at x = 1 and x = 4. Similarly, P(x) - 2R(x) has roots at x = 2 and x = 3.

But maybe this isn't helping me directly. Let me think again. Let's try to write equations using the information we have.

We have P(1) = A + B, P(4) = 4A + B, P(2) = 4A + 2B, P(3) = 6A + 2B. Also, P(0) = 5.

But since P(x) is a cubic polynomial, we can write down its general form. Let me try that. Let me assume:

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

Given that P(0) = c = 5. Therefore, c = 5.

Therefore, P(x) = x^3 + ax^2 + bx + 5.

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

Compute P(1) = 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 earlier, we have:

P(1) = A + B

P(4) = 4A + B

P(2) = 4A + 2B

P(3) = 6A + 2B

So let's write equations:

From P(1) = A + B = a + b + 6    (Equation 1)

From P(4) = 4A + B = 16a + 4b + 69   (Equation 2)

From P(2) = 4A + 2B = 4a + 2b + 13    (Equation 3)

From P(3) = 6A + 2B = 9a + 3b + 32    (Equation 4)

So now we have four equations with variables A, B, a, b. However, we can solve for A and B first using Equations 1, 2, 3, 4, and then relate them to a and b.

Alternatively, since Equations 1, 2, 3, 4 might be redundant. Let me check.

First, let me note that Equations 1 and 3 can be used to solve for A and B in terms of a and b. Similarly, Equations 1 and 2, or Equations 3 and 4. Let me try with Equations 1 and 3 first.

From Equation 1: A + B = a + b + 6

From Equation 3: 4A + 2B = 4a + 2b + 13

Let me write Equation 3 as 2*(2A + B) = 2*(2a + b) + 13? Wait, maybe better to solve the equations:

Let me write:

Equation 1: A + B = a + b + 6

Equation 3: 4A + 2B = 4a + 2b + 13

Let me denote Equation 1 multiplied by 2: 2A + 2B = 2a + 2b + 12

Subtract this from Equation 3:

(4A + 2B) - (2A + 2B) = (4a + 2b + 13) - (2a + 2b + 12)

Which gives:

2A = 2a + 1

Therefore, A = a + 0.5

Then, substitute back into Equation 1: A + B = a + b + 6 => (a + 0.5) + B = a + b + 6 => B = b + 6 - 0.5 => B = b + 5.5

Therefore, A = a + 1/2, B = b + 11/2.

Now, let's check if these hold with Equation 4. Equation 4 is:

6A + 2B = 9a + 3b + 32

Substitute A and B:

6*(a + 1/2) + 2*(b + 11/2) = 9a + 3b + 32

Compute left side:

6a + 3 + 2b + 11 = 6a + 2b + 14

Set equal to right side:

6a + 2b + 14 = 9a + 3b + 32

Bring all terms to left:

6a + 2b +14 -9a -3b -32 = 0 => -3a - b -18 = 0 => 3a + b = -18 (Equation 5)

Similarly, check Equation 2: 4A + B = 16a + 4b + 69

Substitute A and B:

4*(a + 1/2) + (b + 11/2) = 4a + 2 + b + 5.5 = 4a + b + 7.5

Set equal to 16a + 4b + 69:

4a + b + 7.5 = 16a + 4b + 69

Bring all terms to left:

4a + b + 7.5 -16a -4b -69 = 0 => -12a -3b -61.5 = 0 => Multiply both sides by 2 to eliminate decimal:

-24a -6b -123 = 0 => 24a + 6b = -123 => Divide both sides by 3: 8a + 2b = -41 (Equation 6)

So now we have two equations:

Equation 5: 3a + b = -18

Equation 6: 8a + 2b = -41

Let me solve these two equations. Let me use Equation 5 to express b = -18 - 3a. Then substitute into Equation 6.

Equation 6: 8a + 2*(-18 - 3a) = -41

Compute:

8a - 36 - 6a = -41 => 2a - 36 = -41 => 2a = -41 + 36 => 2a = -5 => a = -5/2

Then, from Equation 5: 3*(-5/2) + b = -18 => -15/2 + b = -18 => b = -18 + 15/2 = (-36/2 + 15/2) = -21/2

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

Therefore, we have found a and b. Now, we can find A and B from earlier:

A = a + 1/2 = (-5/2) + 1/2 = (-4/2) = -2

B = b + 11/2 = (-21/2) + 11/2 = (-10/2) = -5

So, R(x) = Ax + B = -2x -5

Therefore, R(x) = -2x -5.

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

For example, P(1) = A + B = -2*1 -5 = -7. Let me compute P(1) using the polynomial:

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

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

Similarly, check P(2):

P(2) = 8 - (5/2)(4) - (21/2)(2) + 5 = 8 - 10 - 21 + 5 = (8 + 5) - (10 + 21) = 13 - 31 = -18.

But according to earlier, P(2) should be 4A + 2B = 4*(-2) + 2*(-5) = -8 -10 = -18. Correct.

Similarly, check P(4):

P(4) = 64 - (5/2)(16) - (21/2)(4) + 5 = 64 - 40 - 42 + 5 = (64 + 5) - (40 + 42) = 69 - 82 = -13.

According to Equation 2: 4A + B = 4*(-2) + (-5) = -8 -5 = -13. Correct.

Also check P(3):

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

According to Equation 4: 6A + 2B = 6*(-2) + 2*(-5) = -12 -10 = -22. Correct.

So, all equations check out. Therefore, our found values for a, b, A, B are correct.

So now, the polynomial P(x) is:

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

But perhaps we can write this with fractions combined:

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

Alternatively, to make it easier for computation, maybe multiply through by 2 to eliminate denominators:

But since we need to find P(5), maybe we can compute it directly.

Alternatively, maybe we can write P(x) in terms of its roots or something else? Not sure. Let me compute P(5) directly.

Compute P(5):

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

Compute each term:

5^3 = 125

(5/2)(5)^2 = (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) + 5 = 10 + 5 = 15.

Wait, so is P(5) equal to 15? Let me check the calculation again step by step to be sure.

First term: 5^3 = 125

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

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

Fourth term: +5

Therefore, adding all together:

125 - 125/2 - 105/2 + 5

Convert 125 and 5 to halves:

125 = 250/2, and 5 = 10/2

Therefore:

250/2 - 125/2 - 105/2 + 10/2 = (250 - 125 - 105 + 10)/2 = (250 - 230 + 10)/2 = (20)/2 = 10.

Wait, hold on. Wait, 250 - 125 is 125, then 125 - 105 is 20, then 20 + 10 is 30? Wait, no:

Wait, let me recast:

Original terms:

125 - 125/2 - 105/2 + 5

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

= 130 - (230/2)

= 130 - 115 = 15

Yes, so earlier step where I converted to halves:

Original:

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

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

= 130 - (230/2)

= 130 - 115 = 15.

Yes, so my first calculation was correct. So P(5) = 15.

But let me double-check by plugging into the polynomial:

P(x) = x³ - (5/2)x² - (21/2)x + 5

So P(5) = 125 - (5/2)(25) - (21/2)(5) + 5 = 125 - 125/2 - 105/2 + 5.

Compute 125 - 125/2 is 125*(1 - 1/2) = 125*(1/2) = 62.5. Then subtract 105/2 which is 52.5. So 62.5 - 52.5 = 10. Then add 5: 10 + 5 = 15. Yep, same result. So P(5) is 15.

Therefore, the answer should be 15.

But wait, let me check again if there's another way to approach this problem, maybe using interpolation or something else, just to confirm.

Alternatively, since we know that R(x) = -2x -5, then when P(x) is divided by (x -1)(x -4), remainder is -2x -5. Also, since the remainder when divided by (x -1)(x -4) is R(x), then P(x) is equal to (x -1)(x -4)Q(x) + R(x). Similarly for the other division. So, since we also have P(0) = 5, which we used to find c = 5, which helped us find the coefficients a and b.

Alternatively, since we have R(x) = -2x -5, and we also have that P(x) - R(x) is divisible by (x -1)(x -4). Therefore, P(x) - (-2x -5) = P(x) + 2x +5 is divisible by (x -1)(x -4). Similarly, P(x) - 2R(x) = P(x) - 2*(-2x -5) = P(x) +4x +10 is divisible by (x -2)(x -3).

Therefore, we can write:

P(x) + 2x +5 = (x -1)(x -4)(some linear polynomial)

Similarly,

P(x) +4x +10 = (x -2)(x -3)(some linear polynomial)

Since P(x) is cubic, then those "some linear polynomials" would be linear. Let me denote:

Let me call the first one: P(x) + 2x +5 = (x -1)(x -4)(mx + n)

Similarly, P(x) +4x +10 = (x -2)(x -3)(px + q)

But since P(x) is same in both, we can set up equations. However, since we already found P(x), maybe this is redundant. But if we didn't know P(x), maybe this could be another way.

Alternatively, let me try plugging x = 1 into P(x) + 2x +5. Since (x -1) divides P(x) +2x +5, then at x = 1, P(1) + 2*1 +5 = 0. Wait, no, since P(x) + 2x +5 is divisible by (x -1), so P(1) + 2*1 +5 should be zero? Wait, no. If P(x) + 2x +5 is divisible by (x -1), then when x = 1, P(1) + 2*1 +5 = 0.

But we know that P(1) = R(1) = -2*1 -5 = -7. Then:

-7 + 2 + 5 = -7 +7 = 0. Correct. Similarly, at x = 4, P(4) + 2*4 +5 = 0. P(4) = R(4) = -8 -5 = -13. Then -13 + 8 +5 = 0. Correct.

Similarly, for the other equation, P(x) +4x +10 is divisible by (x -2)(x -3). So at x = 2, P(2) +4*2 +10 = 0. P(2) = -18. Then -18 + 8 +10 = 0. Correct. Similarly at x = 3, P(3) +4*3 +10 = -22 + 12 +10 = 0. Correct.

So that checks out. Therefore, this method also confirms our previous calculations.

Alternatively, if I didn't have P(x), maybe I could set up equations for the coefficients. But since we already have P(x), and verified all the conditions, and found P(5) = 15, which seems correct.

Wait, but let me check once more if there is a mistake in calculating the coefficients.

We had:

From Equation 5: 3a + b = -18

From Equation 6: 8a + 2b = -41

We solved it by substituting b = -18 - 3a into Equation 6:

8a + 2*(-18 -3a) = 8a -36 -6a = 2a -36 = -41

So 2a = -5 => a = -5/2. Then b = -18 - 3*(-5/2) = -18 + 15/2 = -36/2 + 15/2 = -21/2. Correct.

Therefore, the coefficients are correct. Then the polynomial is:

x^3 - (5/2)x^2 - (21/2)x + 5.

So, if I want to write this polynomial with integer coefficients, I can multiply by 2:

2P(x) = 2x^3 -5x^2 -21x + 10

But maybe not necessary here.

Alternatively, maybe I can find P(5) using another approach. For example, since we know that P(x) - R(x) is divisible by (x -1)(x -4), so P(x) - R(x) = (x -1)(x -4)(something). Similarly, P(x) - 2R(x) = (x -2)(x -3)(something else).

But maybe we can write that P(x) - R(x) = (x -1)(x -4)(kx + m). Similarly, P(x) - 2R(x) = (x -2)(x -3)(lx + n). Since P(x) is cubic, then kx + m and lx + n are linear polynomials.

Alternatively, since we already have R(x) = -2x -5, then:

P(x) = (x -1)(x -4)(kx + m) -2x -5

But also, since P(x) is monic, the leading term of P(x) is x^3. Let me compute the leading term of the right-hand side:

(x -1)(x -4)(kx + m) = (x^2 -5x +4)(kx + m) = kx^3 + (m -5k)x^2 + (4k -5m)x + 4m

Therefore, when we subtract 2x +5, the leading term is kx^3. Since P(x) is monic, so k = 1. So:

P(x) = (x^2 -5x +4)(x + m) -2x -5

Let me expand this:

First multiply (x^2 -5x +4)(x + m):

= x^3 + m x^2 -5x^2 -5m x +4x +4m

= x^3 + (m -5)x^2 + (-5m +4)x + 4m

Then subtract 2x +5:

= x^3 + (m -5)x^2 + (-5m +4)x + 4m -2x -5

= x^3 + (m -5)x^2 + (-5m +4 -2)x + (4m -5)

= x^3 + (m -5)x^2 + (-5m + 2)x + (4m -5)

But we also have P(x) = x^3 - (5/2)x^2 - (21/2)x + 5. Therefore, equate coefficients:

For x^3: 1 = 1. Okay.

For x^2: m -5 = -5/2 => m = -5/2 +5 = 5/2.

For x term: -5m + 2 = -21/2. Let's check with m = 5/2:

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

For constant term: 4m -5 = 5. Let's check:

4*(5/2) -5 = 10 -5 = 5. Correct.

Therefore, m = 5/2. So that's consistent. Therefore, this approach also confirms that. Therefore, P(x) is indeed as we found earlier.

Therefore, P(5) = 15 is correct.

Alternatively, maybe there's a way to use the Chinese Remainder Theorem or something? But since we have two different remainders for two different divisors, but the problem is already solved through the above methods.

Alternatively, maybe use the fact that P(x) - R(x) has roots at 1 and 4, so P(1) = R(1) and P(4) = R(4), which we used.

But regardless, all checks confirm that P(5) = 15. Therefore, I think that's the correct answer.

**Final Answer**
\boxed{15}

We are given a **monic cubic polynomial** $ P(x) $ such that:

- 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) $.
- We are also told that $ P(0) = 5 $.
- We are to find $ P(5) $.

---

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

Let $ P(x) $ be a monic cubic polynomial:
$$
P(x) = x^3 + ax^2 + bx + c
$$

From the problem, we know:
$$
P(0) = c = 5
$$

So we can write:
$$
P(x) = x^3 + ax^2 + bx + 5
$$

---

### Step 2: Use the Remainder Theorem

From the Remainder Theorem, if a polynomial $ P(x) $ is divided by a factor $ (x - a) $, the remainder is simply $ P(a) $.

Since:
- $ P(x) $ mod $ (x - 1)(x - 4) $ gives remainder $ R(x) = Ax + B $,
- $ P(x) $ mod $ (x - 2)(x - 3) $ gives remainder $ 2R(x) = 2Ax + 2B $,

We can deduce:
$$
P(1) = R(1) = A + B \\
P(2) = 2R(2) = 2(2A + B) = 4A + 2B \\
P(3) = 2R(3) = 2(3A + B) = 6A + 2B \\
P(4) = R(4) = 4A + B
$$

So we have the following system of equations:

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

But we can also solve this using the fact that $ P(x) - R(x) $ is divisible by $ (x - 1)(x - 4) $, and $ P(x) - 2R(x) $ is divisible by $ (x - 2)(x - 3) $. That is:

- $ P(x) - R(x) = (x - 1)(x - 4)(mx + n) $
- $ P(x) - 2R(x) = (x - 2)(x - 3)(px + q) $

However, a more direct and simpler path is to solve the above system of four equations for $ A, B, a, b $, using substitution and elimination.

---

### Step 3: Solve the System of Equations

From equation (1):
$$
A + B = a + b + 6 \quad \text{(Equation 1)}
$$

From equation (3):
$$
4A + 2B = 4a + 2b + 13 \quad \text{(Equation 3)}
$$

Multiply Equation 1 by 2:
$$
2A + 2B = 2a + 2b + 12
$$

Subtract from Equation 3:
$$
(4A + 2B) - (2A + 2B) = (4a + 2b + 13) - (2a + 2b + 12)
$$
$$
2A = 2a + 1 \Rightarrow A = a + \frac{1}{2}
$$

Substitute into (1):
$$
A + B = a + b + 6 \Rightarrow a + \frac{1}{2} + B = a + b + 6 \Rightarrow B = b + 5.5
$$

Now use equation (4):
$$
6A + 2B = 9a + 3b + 32
$$

Substitute $ A = a + \frac{1}{2} $, $ B = b + \frac{11}{2} $:
$$
6(a + 0.5) + 2(b + 5.5) = 9a + 3b + 32
$$
$$
6a + 3 + 2b + 11 = 9a + 3b + 32
\Rightarrow 6a + 2b + 14 = 9a + 3b + 32
$$
$$
-3a - b = -18 \Rightarrow 3a + b = -18 \quad \text{(Equation 5)}
$$

Now consider equation (2):
$$
4A + B = 16a + 4b + 69
$$
Substitute $ A = a + \frac{1}{2} $, $ B = b + \frac{11}{2} $:
$$
4(a + 0.5) + b + \frac{11}{2} = 16a + 4b + 69
\Rightarrow 4a + 2 + b + 5.5 = 16a + 4b + 69
\Rightarrow 4a + b + 7.5 = 16a + 4b + 69
\Rightarrow -12a -3b = -61.5
\Rightarrow 8a + 2b = -41 \quad \text{(Equation 6)}
$$

Now solve the system:
$$
\begin{cases}
3a + b = -18 \\
8a + 2b = -41
\end{cases}
$$

From first equation:
$$
b = -18 - 3a
$$

Substitute into second:
$$
8a + 2(-18 - 3a) = -41 \Rightarrow 8a - 36 - 6a = -41 \Rightarrow 2a = -5 \Rightarrow a = -\frac{5}{2}
$$
$$
b = -18 - 3(-\frac{5}{2}) = -18 + \frac{15}{2} = -\frac{21}{2}
$$

So the polynomial becomes:
$$
P(x) = x^3 - \frac{5}{2}x^2 - \frac{21}{2}x + 5
$$

---

### Step 4: Compute $ P(5) $

$$
P(5) = 5^3 - \frac{5}{2}(5)^2 - \frac{21}{2}(5) + 5
= 125 - \frac{125}{2} - \frac{105}{2} + 5
= 125 - \frac{230}{2} + 5
= 125 - 115 + 5 = \boxed{15}
$$

---

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