## College Algebra (10th Edition)

The solution set is $\left\{-1, -\frac{2}{3}\right\}$.
RECALL: A quadratic trinomial $ax^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = ac$ and $d+e =b$ If such integers exist, then $ax^2+bx + c = ax^2+dx + ex + c$, and may be factored by grouping. The trinomial in the equation has $a=3$, $b=5$, and $c=2$. Thus, $ac = 3(2) = 6$ The factors of $6$ whose sum is equal to the coefficient of the middle term (which is $b=5$) are $3$ and $2$. This means that $d=3$ and $e=2$. Rewrite the middle term of the trinomial as $3x$ +$2x$ to obtain: $3x^2+5x+2 = 0 \\3x^2+3x+2x+2 = 0$ Regroup then factor out the GCF in each group to obtain: $(3x^2+3x)+(2x+2)=0 \\3x(x+1) +2(x+1)=0$ Factor out $x+1$ to obtain: $(x+1)(3x+2)=0$ Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain: $\begin{array}{ccc} &x+1=0 &\text{ or } &3x+2=0 \\&x=-1 &\text{ or } &3x=-2 \\&x=-1 &\text{ or } &x=-\frac{2}{3} \end{array}$ Therefore the solution set is $\left\{-1, -\frac{2}{3}\right\}$.