## College Algebra (6th Edition)

solution set: $\{ -3,\ \displaystyle \frac{1}{2},\ 2 \}$
The graph suggests that x=$-3$ is a solution to the equation. We check by synthetic division(the last number in the last row is $f(-3)$ $( 2x^{3}+x^{2}-13x+6)\div(x+3)$ $\begin{array}{lllllll} \underline{-3}| & 2 & 1 & -13 & 6 & & \\ & & -6 & 15 & -6 & & \\ & -- & -- & -- & -- & -- & \\ & 2 & -5 & 2 & |\overline{ 0 } & & \end{array}$ By the Factor Theorem, $-3$ is a zero, $(x+3)$ is a factor of $f(x)$. $f(x)=(x+3)(2x^{2}-5x+2)$ we try to factor the trinomial by searching for two integers whose product is $2\cdot 2=4$ and whose sum is $-5$.... We find $-1$ and $-4:\quad$ $2x^{2}-5x+2=2x^{2}-4x-x+2\quad$...factor in pairs $=2x(x-2)-(x-2) = (2x-1)(x-2)$ ... factor out 2 from the first... $=2(x-\displaystyle \frac{1}{2})(x-2)$ So, f(x) has three factors of form $(x-c)$, $f(x)=2(x-\displaystyle \frac{1}{2})(x-2)(x+3)$ and by (b) of the Factor Theorem, zeros of f: $\displaystyle \frac{1}{2},\ 2,\ -3$ make up the solution set of the equation $f(x)=0:$ $\{ -3,\ \displaystyle \frac{1}{2},\ 2 \}$