Question

Solve the polynomial equation p(x)=x3+2x2−5x−6
𝑝
(
𝑥
)
=
𝑥
3
+
2
𝑥
2
−
5
𝑥
−
6
. Which of the following is a factor?(1 point)
Responses

x+1
𝑥
+
1
x plus 1

x+2
𝑥
+
2
x plus 2

x−1
𝑥
−
1
x minus 1

x−3

Answer 1

To determine which of the provided options is a factor of the polynomial \( p(x) = x^3 + 2x^2 - 5x - 6 \), we can apply the Factor Theorem. The Factor Theorem states that if \( p(c) = 0 \), then \( x - c \) is a factor of \( p(x) \).

We will evaluate \( p(x) \) at each of the possible roots:

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

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

3. **For \( x = -1 \)** (already calculated):
\[
p(-1) = 0
\]

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

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

Based on these calculations, the only factor of \( p(x) = x^3 + 2x^2 - 5x - 6 \) among the options provided is \( x + 1 \).