Math worksheet for practicing the Remainder and Factor Theorems with polynomial division and synthetic division exercises.

Problem Overview:

The worksheet involves two main sections:
1. Using the Remainder Theorem to find the remainder when a polynomial is divided by a binomial and determining if the binomial is a factor.
2. Factoring polynomials completely using synthetic division, given one zero.

We will solve each part step by step.

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Section 1: Using the Remainder Theorem

The Remainder Theorem states that if a polynomial \( f(x) \) is divided by \( (x - c) \), the remainder is \( f(c) \). If the remainder is 0, then \( (x - c) \) is a factor of \( f(x) \).

#### Problem 1:
\[
(x^3 + x^2 - 10) \div (x + 3)
\]
- Here, the divisor is \( x + 3 \), so \( c = -3 \).
- Evaluate \( f(-3) \) where \( f(x) = x^3 + x^2 - 10 \):
\[
f(-3) = (-3)^3 + (-3)^2 - 10 = -27 + 9 - 10 = -28
\]
- The remainder is \( -28 \).
- Since the remainder is not 0, \( x + 3 \) is not a factor of \( x^3 + x^2 - 10 \).

#### Problem 2:
\[
(2x^3 - 3x^2 - 10x + 3) \div (x - 3)
\]
- Here, the divisor is \( x - 3 \), so \( c = 3 \).
- Evaluate \( f(3) \) where \( f(x) = 2x^3 - 3x^2 - 10x + 3 \):
\[
f(3) = 2(3)^3 - 3(3)^2 - 10(3) + 3 = 2(27) - 3(9) - 30 + 3 = 54 - 27 - 30 + 3 = 0
\]
- The remainder is \( 0 \).
- Since the remainder is 0, \( x - 3 \) is a factor of \( 2x^3 - 3x^2 - 10x + 3 \).

#### Problem 3:
\[
(x^4 + 5x^3 - 14x^2) \div (x - 2)
\]
- Here, the divisor is \( x - 2 \), so \( c = 2 \).
- Evaluate \( f(2) \) where \( f(x) = x^4 + 5x^3 - 14x^2 \):
\[
f(2) = (2)^4 + 5(2)^3 - 14(2)^2 = 16 + 5(8) - 14(4) = 16 + 40 - 56 = 0
\]
- The remainder is \( 0 \).
- Since the remainder is 0, \( x - 2 \) is a factor of \( x^4 + 5x^3 - 14x^2 \).

#### Problem 4:
\[
(2x^4 + 4x^3 - x^2 + 9) \div (x + 1)
\]
- Here, the divisor is \( x + 1 \), so \( c = -1 \).
- Evaluate \( f(-1) \) where \( f(x) = 2x^4 + 4x^3 - x^2 + 9 \):
\[
f(-1) = 2(-1)^4 + 4(-1)^3 - (-1)^2 + 9 = 2(1) + 4(-1) - 1 + 9 = 2 - 4 - 1 + 9 = 6
\]
- The remainder is \( 6 \).
- Since the remainder is not 0, \( x + 1 \) is not a factor of \( 2x^4 + 4x^3 - x^2 + 9 \).

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Section 2: Factoring Polynomials Using Synthetic Division

#### Problem 5:
\[
f(x) = x^3 - 8x^2 + 19x - 12; \quad \text{Given zero: } 3
\]
- Use synthetic division with \( c = 3 \):
\[
\begin{array}{r|rrrr}
3 & 1 & -8 & 19 & -12 \\
& & 3 & -15 & 12 \\
\hline
& 1 & -5 & 4 & 0 \\
\end{array}
\]
- The quotient is \( x^2 - 5x + 4 \).
- Factor \( x^2 - 5x + 4 \):
\[
x^2 - 5x + 4 = (x - 1)(x - 4)
\]
- Therefore, the complete factorization is:
\[
f(x) = (x - 3)(x - 1)(x - 4)
\]

#### Problem 6:
\[
f(x) = x^4 + 2x^3 - 13x^2 + 10x; \quad \text{Given zero: } -5
\]
- Use synthetic division with \( c = -5 \):
\[
\begin{array}{r|rrrrr}
-5 & 1 & 2 & -13 & 10 & 0 \\
& & -5 & 15 & -10 & 0 \\
\hline
& 1 & -3 & 2 & 0 & 0 \\
\end{array}
\]
- The quotient is \( x^3 - 3x^2 + 2x \).
- Factor \( x^3 - 3x^2 + 2x \):
\[
x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x - 1)(x - 2)
\]
- Therefore, the complete factorization is:
\[
f(x) = (x + 5)x(x - 1)(x - 2)
\]

#### Problem 7:
\[
f(x) = x^3 + 6x^2 + 11x + 6; \quad \text{Given zero: } -2
\]
- Use synthetic division with \( c = -2 \):
\[
\begin{array}{r|rrrr}
-2 & 1 & 6 & 11 & 6 \\
& & -2 & -8 & -6 \\
\hline
& 1 & 4 & 3 & 0 \\
\end{array}
\]
- The quotient is \( x^2 + 4x + 3 \).
- Factor \( x^2 + 4x + 3 \):
\[
x^2 + 4x + 3 = (x + 1)(x + 3)
\]
- Therefore, the complete factorization is:
\[
f(x) = (x + 2)(x + 1)(x + 3)
\]

#### Problem 8:
\[
f(x) = x^4 - 2x^3 - 5x^2 + 6x; \quad \text{Given zero: } -2
\]
- Use synthetic division with \( c = -2 \):
\[
\begin{array}{r|rrrrr}
-2 & 1 & -2 & -5 & 6 & 0 \\
& & -2 & 8 & -6 & 0 \\
\hline
& 1 & -4 & 3 & 0 & 0 \\
\end{array}
\]
- The quotient is \( x^3 - 4x^2 + 3x \).
- Factor \( x^3 - 4x^2 + 3x \):
\[
x^3 - 4x^2 + 3x = x(x^2 - 4x + 3) = x(x - 1)(x - 3)
\]
- Therefore, the complete factorization is:
\[
f(x) = (x + 2)x(x - 1)(x - 3)
\]

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Final Answers:

1. Remainder: \(-28\), Not a factor.
2. Remainder: \(0\), Factor.
3. Remainder: \(0\), Factor.
4. Remainder: \(6\), Not a factor.
5. Factorization: \((x - 3)(x - 1)(x - 4)\).
6. Factorization: \((x + 5)x(x - 1)(x - 2)\).
7. Factorization: \((x + 2)(x + 1)(x + 3)\).
8. Factorization: \((x + 2)x(x - 1)(x - 3)\).

\[
\boxed{
\begin{array}{l}
1. -28, \text{Not a factor} \\
2. 0, \text{Factor} \\
3. 0, \text{Factor} \\
4. 6, \text{Not a factor} \\
5. (x - 3)(x - 1)(x - 4) \\
6. (x + 5)x(x - 1)(x - 2) \\
7. (x + 2)(x + 1)(x + 3) \\
8. (x + 2)x(x - 1)(x - 3)
\end{array}
}
\]