Remainder Theorem and Factor Theorem interactive worksheet | Live ... - Free Printable

Problem Overview:

The task involves using the Remainder Theorem to determine the remainder when a polynomial \( f(x) \) is divided by a linear divisor of the form \( x - c \). The Remainder Theorem states that if a polynomial \( f(x) \) is divided by \( x - c \), the remainder is \( f(c) \). Additionally, the Factor Theorem states that \( x - c \) is a factor of \( f(x) \) if and only if \( f(c) = 0 \).

We will solve each part step by step.

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Step-by-Step Solution:

#### 1. \( x^4 - 2x^3 + x^2 - 14x + 6 \) by \( x - 3 \)

- Polynomial: \( f(x) = x^4 - 2x^3 + x^2 - 14x + 6 \)
- Divisor: \( x - 3 \)
- Using the Remainder Theorem: The remainder when \( f(x) \) is divided by \( x - 3 \) is \( f(3) \).

\[
f(3) = (3)^4 - 2(3)^3 + (3)^2 - 14(3) + 6
\]

- Calculate each term:
\[
(3)^4 = 81, \quad -2(3)^3 = -2 \cdot 27 = -54, \quad (3)^2 = 9, \quad -14(3) = -42, \quad 6 = 6
\]

- Sum the terms:
\[
f(3) = 81 - 54 + 9 - 42 + 6 = 0
\]

- Remainder: \( 0 \)

- Conclusion: Since the remainder is \( 0 \), \( x - 3 \) is a factor of \( x^4 - 2x^3 + x^2 - 14x + 6 \).

#### 2. \( x^4 - 6x^2 - 13x + 13 \) by \( x - 3 \)

- Polynomial: \( f(x) = x^4 - 6x^2 - 13x + 13 \)
- Divisor: \( x - 3 \)
- Using the Remainder Theorem: The remainder when \( f(x) \) is divided by \( x - 3 \) is \( f(3) \).

\[
f(3) = (3)^4 - 6(3)^2 - 13(3) + 13
\]

- Calculate each term:
\[
(3)^4 = 81, \quad -6(3)^2 = -6 \cdot 9 = -54, \quad -13(3) = -39, \quad 13 = 13
\]

- Sum the terms:
\[
f(3) = 81 - 54 - 39 + 13 = 1
\]

- Remainder: \( 1 \)

- Conclusion: Since the remainder is not \( 0 \), \( x - 3 \) is not a factor of \( x^4 - 6x^2 - 13x + 13 \).

#### 3. \( x^3 + 5x^2 + 4x + 16 \) by \( x + 5 \)

- Polynomial: \( f(x) = x^3 + 5x^2 + 4x + 16 \)
- Divisor: \( x + 5 \)
- Using the Remainder Theorem: The remainder when \( f(x) \) is divided by \( x + 5 \) is \( f(-5) \).

\[
f(-5) = (-5)^3 + 5(-5)^2 + 4(-5) + 16
\]

- Calculate each term:
\[
(-5)^3 = -125, \quad 5(-5)^2 = 5 \cdot 25 = 125, \quad 4(-5) = -20, \quad 16 = 16
\]

- Sum the terms:
\[
f(-5) = -125 + 125 - 20 + 16 = 1
\]

- Remainder: \( 1 \)

- Conclusion: Since the remainder is not \( 0 \), \( x + 5 \) is not a factor of \( x^3 + 5x^2 + 4x + 16 \).

#### 4. \( x^3 - x^2 - 14x + 24 \) by \( x + 4 \)

- Polynomial: \( f(x) = x^3 - x^2 - 14x + 24 \)
- Divisor: \( x + 4 \)
- Using the Remainder Theorem: The remainder when \( f(x) \) is divided by \( x + 4 \) is \( f(-4) \).

\[
f(-4) = (-4)^3 - (-4)^2 - 14(-4) + 24
\]

- Calculate each term:
\[
(-4)^3 = -64, \quad -(-4)^2 = -16, \quad -14(-4) = 56, \quad 24 = 24
\]

- Sum the terms:
\[
f(-4) = -64 - 16 + 56 + 24 = 0
\]

- Remainder: \( 0 \)

- Conclusion: Since the remainder is \( 0 \), \( x + 4 \) is a factor of \( x^3 - x^2 - 14x + 24 \).

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Final Answers:

1. Remainder: \( 0 \), \( x - 3 \) is a factor.
2. Remainder: \( 1 \), \( x - 3 \) is not a factor.
3. Remainder: \( 1 \), \( x + 5 \) is not a factor.
4. Remainder: \( 0 \), \( x + 4 \) is a factor.

\[
\boxed{
\begin{array}{ll}
1. & \text{Remainder: } 0, \text{ Factor: Yes} \\
2. & \text{Remainder: } 1, \text{ Factor: No} \\
3. & \text{Remainder: } 1, \text{ Factor: No} \\
4. & \text{Remainder: } 0, \text{ Factor: Yes} \\
\end{array}
}
\]