Question

the fundamental theorem of algebra quick check
solve the polynomial equation $p(x) = x^3 + 2x^2 - 5x - 6$. which of the following is a factor? (1 point)
$\circ$ $x + 1$
$\circ$ $x + 2$
$\circ$ $x - 3$
$\circ$ $x - 1$

Explanation:

Step1: Recall Factor Theorem

If \(x - a\) is a factor of \(p(x)\), then \(p(a)=0\). For a factor \(x + b\) (which is \(x - (-b)\)), we check \(p(-b)=0\).

Step2: Check Option \(x + 1\) (i.e., \(a=-1\))

Substitute \(x=-1\) into \(p(x)=x^{3}+2x^{2}-5x - 6\):
\(p(-1)=(-1)^{3}+2(-1)^{2}-5(-1)-6=-1 + 2 + 5 - 6=0\).

Step3: Verify Other Options (Optional)

• For \(x + 2\) (\(x=-2\)): \(p(-2)=(-2)^{3}+2(-2)^{2}-5(-2)-6=-8 + 8 + 10 - 6=4

eq0\).

• For \(x - 3\) (\(x=3\)): \(p(3)=3^{3}+2(3)^{2}-5(3)-6=27 + 18 - 15 - 6=24

eq0\).

• For \(x - 1\) (\(x=1\)): \(p(1)=1^{3}+2(1)^{2}-5(1)-6=1 + 2 - 5 - 6=-8

eq0\).

Answer:

A. \(x + 1\)