Worksheet for practicing the Remainder Theorem and Factor Theorem with polynomial division exercises.

Problem Analysis:

The task involves using the Remainder Theorem to find the remainder when a polynomial is divided by a linear divisor. The Remainder Theorem states that if a polynomial \( P(x) \) is divided by \( x - c \), the remainder is \( P(c) \). Additionally, we need to determine if the divisor is a factor of the polynomial. If \( P(c) = 0 \), then \( x - c \) is a factor of \( P(x) \).

Let's solve each part step by step.

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Problem 1:

Find the remainder when \( x^4 + 4x^3 - 11x - 5 \) is divided by \( x + 3 \).

#### Step 1: Identify the divisor and rewrite it in the form \( x - c \).
The divisor is \( x + 3 \), which can be rewritten as \( x - (-3) \). Thus, \( c = -3 \).

#### Step 2: Apply the Remainder Theorem.
We need to evaluate \( P(-3) \) for the polynomial \( P(x) = x^4 + 4x^3 - 11x - 5 \).

\[
P(-3) = (-3)^4 + 4(-3)^3 - 11(-3) - 5
\]

#### Step 3: Calculate each term.
- \( (-3)^4 = 81 \)
- \( 4(-3)^3 = 4(-27) = -108 \)
- \( -11(-3) = 33 \)
- The constant term is \( -5 \)

So,
\[
P(-3) = 81 - 108 + 33 - 5
\]

#### Step 4: Simplify the expression.
\[
P(-3) = 81 - 108 + 33 - 5 = 81 + 33 - 108 - 5 = 114 - 113 = 1
\]

#### Step 5: Determine if the divisor is a factor.
Since \( P(-3) = 1 \neq 0 \), \( x + 3 \) is not a factor of \( P(x) \).

#### Final Answer for Problem 1:
\[
\boxed{1} \quad \text{(No, it is not a factor)}
\]

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Problem 2:

Find the remainder when \( x^3 + 5x^2 - 2x - 24 \) is divided by \( x - 2 \).

#### Step 1: Identify the divisor and rewrite it in the form \( x - c \).
The divisor is \( x - 2 \). Thus, \( c = 2 \).

#### Step 2: Apply the Remainder Theorem.
We need to evaluate \( P(2) \) for the polynomial \( P(x) = x^3 + 5x^2 - 2x - 24 \).

\[
P(2) = (2)^3 + 5(2)^2 - 2(2) - 24
\]

#### Step 3: Calculate each term.
- \( (2)^3 = 8 \)
- \( 5(2)^2 = 5(4) = 20 \)
- \( -2(2) = -4 \)
- The constant term is \( -24 \)

So,
\[
P(2) = 8 + 20 - 4 - 24
\]

#### Step 4: Simplify the expression.
\[
P(2) = 8 + 20 - 4 - 24 = 28 - 28 = 0
\]

#### Step 5: Determine if the divisor is a factor.
Since \( P(2) = 0 \), \( x - 2 \) is a factor of \( P(x) \).

#### Final Answer for Problem 2:
\[
\boxed{0} \quad \text{(Yes, it is a factor)}
\]

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Problem 3:

Find the remainder when \( 4x^3 - 12x^2 - x + 3 \) is divided by \( x - 3 \).

#### Step 1: Identify the divisor and rewrite it in the form \( x - c \).
The divisor is \( x - 3 \). Thus, \( c = 3 \).

#### Step 2: Apply the Remainder Theorem.
We need to evaluate \( P(3) \) for the polynomial \( P(x) = 4x^3 - 12x^2 - x + 3 \).

\[
P(3) = 4(3)^3 - 12(3)^2 - (3) + 3
\]

#### Step 3: Calculate each term.
- \( 4(3)^3 = 4(27) = 108 \)
- \( -12(3)^2 = -12(9) = -108 \)
- \( -(3) = -3 \)
- The constant term is \( +3 \)

So,
\[
P(3) = 108 - 108 - 3 + 3
\]

#### Step 4: Simplify the expression.
\[
P(3) = 108 - 108 - 3 + 3 = 0
\]

#### Step 5: Determine if the divisor is a factor.
Since \( P(3) = 0 \), \( x - 3 \) is a factor of \( P(x) \).

#### Final Answer for Problem 3:
\[
\boxed{0} \quad \text{(Yes, it is a factor)}
\]

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Final Answers:

1. \( \boxed{1} \) (No, it is not a factor)
2. \( \boxed{0} \) (Yes, it is a factor)
3. \( \boxed{0} \) (Yes, it is a factor)