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

We can use synthetic division with the root $x=-1$:
$$
\begin{array}{c|cccc}
-1 & 1 & 2 & -1 & -2 \
\hline 
& & -1 & -1 & 2 \
\end{array}
$$
The remainder is 2, which means that the polynomial can be factored as $(x+1)(x^2-x-2)$.
We can factor the quadratic as $(x+1)(x-2)(x+1)$, which gives the complete factorization:
$$(x+1)^2(x-2).$$
The real solutions of the equation are the roots of the factors, which are:
$$x=-1\quad \text{(double root)},\quad x=1,\quad \text{and}\quad x=-2.$$
Therefore, the correct choice is $\boxed{\text{(B)}}$.

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 $x ^ { 3 } + 2 x ^ { 2 } - x - 2 = 0$