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

x=2 is a solution because $f(2)=0$ Solution set = $\{-3,-1,2\}$
The graph suggests that $f(2)=0$. We divide f(x) with $(x-2)$, using synthetic division. $\begin{array}{rrrrrrrr} \underline{2}| & 1 & 2 & -5 &-6 & & & & \\ & & 2 & 8 &6 & & & & \\ & --& -- &-- &-- & & & & \\ & 1 & 4 & 3 & \fbox{0} & & & & \end{array}$ By the remainder theorem, $f(2)=0.$ $\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}\quad$with remainder =0, $x^{3}+2x^{2}-5x-6=(x-2)(x^{2}+4x+3)$ ... the trinomial on the RHS can be factored (two factors of $3$ with sum $4$ are $+1$ and $+3$. $x^{3}+2x^{2}-5x-6=(x-2)(x+1)(x+3)$ Solve $(x-2)(x+1)(x+3)=0\quad$.... apply the zero product principle, Solution set = $\{-3,-1,2\}$