If $x = - 1$ is a root of $x ^ { 3 } + 2 x ^ { 2 } - x - 2 = 0$, use synthetic division to factor the polynomial completely and list all real solutions of the equation.
A) $( x - 2 ) ( x + 1 ) ( x - 1 ) ; x = 2 , - 1,1$
B) $( x + 2 ) ( x + 1 ) ( x - 1 ) ; x = - 2 , - 1,1$
C) $( x + 2 ) ( x + 1 ) ^ { 2 } ; x = - 2 , - 1$
D) $( x + 2 ) ^ { 2 } ( x + 1 ) ; x = - 2 , - 1$
E) $( x + 2 ) ( x - 2 ) ( x - 1 ) ; x = - 2,2,1$

Question

Quizplus · Accepted Answer

Using synthetic division, we get:

\begin{array}{cccc} & & - 1 & 2 & - 1 & - 2 \ \cline{2-6} -1 & \big) & 1 & 2 & - 1 & - 2 \ & & - & - 1 & - 1 & 2 \ \hline & 1 & 1 & - 2 & - 3 & 0 \end{array}

So, we have:

$$
x^3 + 2x^2 - x - 2 = (x+1)(x^2-3x-2) = (x+1)(x+2)(x-1).
$$

Therefore, the real solutions are $x=-2, -1,$ and $x=1$. Option B is the only one that includes all three of these solutions.

If x=−1x = - 1x=−1 Is a Root Of x3+2x2−x−2=0x ^ { 3 } + 2 x ^ { 2 } - x - 2 = 0x3+2x2−x−2=0

If $x = - 1$ Is a Root Of $x ^ { 3 } + 2 x ^ { 2 } - x - 2 = 0$