## College Algebra (10th Edition)

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