## Intermediate Algebra for College Students (7th Edition)

Remainder = 0, Solution set =$\{-1,1,2\}$
Divide $x^{3}-2x^{2}-x+2$ with $x+1,\qquad (c=-1)$ $\begin{array}{rrrrrrrr} \underline{-1}| &1 & -2 & -1 & 2 & & & & \\ & & -1 & 3 &-2 & & & & \\ & --& -- &-- &-- & & & & \\ & 1 & -3 & 2 & \fbox{0} & & & & \end{array}$ $\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$, and the remainder is 0, so $\displaystyle \frac{x^{3}-2x^{2}-x+2}{(x+1)}=x^{2}-3x+2$ ... the trinomial on the RHS can be factored (two factors of $2$ with sum $-3$ are $-1$ and $-2$. Thus, $x^{2}-3x+2 =(x-2)(x-1)\quad$ and $x^{3}-2x^{2}-x+2=(x+1)(x-1)(x-2)$ The equation $x^{3}-2x^{2}-x+2=0$ becomes $(x+1)(x-1)(x-2) =0\quad$ ... apply the zero product principle Solution set =$\{-1,1,2\}$