## College Algebra (6th Edition)

solution set: $\{-2,\ \displaystyle \frac{1}{2},\ 3\}$
The problem, in fact, asks us to find zeros of f. First, let's divide $(2x^{3}-3x^{2}-11x+6)\div(x+2)\qquad (c=-2)$ $\begin{array}{lllllll} \underline{-2}| & 2 & -3 & -11 & 6 & & \\ & & -4 & 14 & -6 & & \\ & -- & -- & -- & -- & -- & \\ & 2 & -7 & 3 & |\overline{ 0 } & & \end{array}$ So, $(x+2)$ is a factor of $f(x).$ $f(x)=(x+2)(2x^{2}-7x+3)$ we try to factor the trinomial by searching for two integers whose product is $2\cdot(3)=6$ and whose sum is $-7$.... We find $-1$ and $-6:\quad$ $2x^{2}-7x+3= 2x^{2}-x-6x+3$ ... factor in pairs ... $=x(2x-1)-3(2x-1)$ $=(2x-1)(x-3)$ ... factor out 2 from the first parentheses ... $=2(x-\displaystyle \frac{1}{2})(x-3)$ So, $f(x)=2(x-\displaystyle \frac{1}{2})(x-3)(x+2)$ and by (b) of the Factor Theorem, the solution set of the equation $f(x)=0$ are the zeros of f: $\{-2,\ \displaystyle \frac{1}{2},\ 3\}$