## College Algebra (6th Edition)

$\{ -3,\ -1,\ 2 \}$
The graph suggests that x=2 is a solution to the equation. We check by synthetic division(the last number in the last row is $f(2)$ Also, if it is a zero, then, by the Factor Theorem, $(x-2)$ is a factor of f... $(x^{3}+2x^{2}-5x-6)\div(x-2)$ $\begin{array}{lllllll} \underline{2}| & 1 & 2 & -5 & -6 & & \\ & & 2 & 8 & 6 & & \\ & -- & -- & -- & -- & -- & \\ & 1 & 4 & 3 & |\overline{ 0 } & & \end{array}$ So, $(x-2)$ is a factor of $f(x).$ $f(x)=(x-2)(x^{2}+4x+3)$ we try to factor the trinomial by searching for two integers whose product is $3$ and whose sum is $4$.... We find $1$ and $3:\quad x^{2}+4x+3=(x+1)(x+3)$ So, f(x) has three factors of form $(x-c)$, $f(x)=(x-2)(x+1)(x+3)$, and by (b) of the Factor Theorem, zeros of f: $-3,\ -1,\ 2$ make up the solution set of the equation $f(x)=0:$ $\{ -3,\ -1,\ 2 \}$