Question

Solve the polynomial equation p(x)=x3+2x2−5x−6 . Which of the following is a factor?(1 point)
Responses

x+1
x plus 1

x−3
x minus 3

x+2
x plus 2

x−1

Answer 1

To determine which of the given options is a factor of the polynomial \( p(x) = x^3 + 2x^2 - 5x - 6 \), we can use the Factor Theorem. This theorem states that \( (x - r) \) is a factor of \( p(x) \) if and only if \( p(r) = 0 \).

We will evaluate \( p(x) \) at the possible roots derived from the answer choices:

1. **For \( x + 1 \) (i.e., \( r = -1 \))**:
\[
p(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0
\]
Since \( p(-1) = 0 \), \( x + 1 \) is a factor of \( p(x) \).

2. **For \( x - 3 \) (i.e., \( r = 3 \))**:
\[
p(3) = 3^3 + 2(3)^2 - 5(3) - 6 = 27 + 18 - 15 - 6 = 24
\]
\( p(3) \neq 0 \), so \( x - 3 \) is not a factor.

3. **For \( x + 2 \) (i.e., \( r = -2 \))**:
\[
p(-2) = (-2)^3 + 2(-2)^2 - 5(-2) - 6 = -8 + 8 + 10 - 6 = 4
\]
\( p(-2) \neq 0 \), so \( x + 2 \) is not a factor.

4. **For \( x - 1 \) (i.e., \( r = 1 \))**:
\[
p(1) = 1^3 + 2(1)^2 - 5(1) - 6 = 1 + 2 - 5 - 6 = -8
\]
\( p(1) \neq 0 \), so \( x - 1 \) is not a factor.

Based on the evaluations, the only factor of the polynomial \( p(x) \) is:

**\( x + 1 \)**.