College Algebra (10th Edition)

The solution set is $\left\{-6, -1\right\}$.
RECALL: A quadratic trinomial $x^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = c$ and $d+e =b$ If such integers exist, then $x^2+bx + c = (x+d)(x+e)$ The trinomial in the given equation has $c=6$ and $b= 7$. Note that $6=6(1)$ and $6+1 = 7$ This means that $d=6$ and $e = 1$ and the given trinomial may be factored as: $v^2+7v+6 \\=(v+d)(v+e) \\=(v+6(v+1)$ Thus, the given equation is equivalent to: $(v+6)(v+1) =0$ Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain: $\begin{array}{ccc} &v+6=0 &\text{ or } &v+1=0 \\&v=-6 &\text{ or } &v=-1 \end{array}$ Therefore the solution set is $\left\{-6, -1\right\}$.