Polynomial division problems using the remainder theorem.

Let’s solve each problem using the Remainder Theorem.

The Remainder Theorem says:
> If you divide a polynomial \( f(x) \) by \( (x - a) \), the remainder is just \( f(a) \).

So for each problem, we’ll plug in the value of \( x \) that makes the divisor zero. For example:
- If dividing by \( (x + 2) \), set \( x = -2 \)
- If dividing by \( (x - 3) \), set \( x = 3 \)
- And so on.

We’ll compute \( f(a) \) for each case — that’s our remainder.

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Problem 1:
\( (4x^3 + 8x^2 - 3x - 6) ÷ (x + 2) \)

Set \( x = -2 \)

Compute:
\[
f(-2) = 4(-2)^3 + 8(-2)^2 - 3(-2) - 6
\]
\[
= 4(-8) + 8(4) + 6 - 6
\]
\[
= -32 + 32 + 6 - 6 = 0
\]

✔ Remainder = 0

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Problem 2:
\( (x^3 - 4x^2 + x - 4) ÷ (x - 3) \)

Set \( x = 3 \)

Compute:
\[
f(3) = (3)^3 - 4(3)^2 + 3 - 4
\]
\[
= 27 - 4(9) + 3 - 4
\]
\[
= 27 - 36 + 3 - 4 = (27 + 3) - (36 + 4) = 30 - 40 = -10
\]

✔ Remainder = -10

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Problem 3:
\( (4x^3 - 20x^2 + 25x - 125) ÷ (x - 6) \)

Set \( x = 6 \)

Compute:
\[
f(6) = 4(6)^3 - 20(6)^2 + 25(6) - 125
\]
\[
= 4(216) - 20(36) + 150 - 125
\]
\[
= 864 - 720 + 150 - 125
\]
Step-by-step:
864 - 720 = 144
144 + 150 = 294
294 - 125 = 169

✔ Remainder = 169

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Problem 4:
\( (x^3 - 3x^2 + 2x - 6) ÷ (x - 2) \)

Set \( x = 2 \)

Compute:
\[
f(2) = (2)^3 - 3(2)^2 + 2(2) - 6
\]
\[
= 8 - 3(4) + 4 - 6
\]
\[
= 8 - 12 + 4 - 6
\]
Step-by-step:
8 - 12 = -4
-4 + 4 = 0
0 - 6 = -6

✔ Remainder = -6

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Problem 5:
\( (4x^3 - 16x^2 + x - 4) ÷ (x - 5) \)

Set \( x = 5 \)

Compute:
\[
f(5) = 4(5)^3 - 16(5)^2 + 5 - 4
\]
\[
= 4(125) - 16(25) + 1
\]
\[
= 500 - 400 + 1 = 101
\]

✔ Remainder = 101

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Final Answer:
0, -10, 169, -6, 101