## Intermediate Algebra: Connecting Concepts through Application

$\color{blue}{\left\{5, -\frac{3}{2}\right\}}$
Add $5$ to both sides to obtain: $2x^2+13x+15=0$ 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 above has $a=2$, $b=13$, and $c=15$. Thus, $ac = 2(15) = 30$ Note that $30=10(3)$ and $10+3=13$ This means that $d=10$ and $e=3$. Rewrite the middle term of the trinomial as $10x$ +$3x$ to obtain: $2x^2+13x+15 =2x^2+10x+3x+15$ Group the first two terms together and the last two terms together. Then, factor out the GCF in each group to obtain: $=(2x^2+10x)+(3x+15) \\=2x(x+5) +3(x+5)$ Factor out the GCF $x+5$ to obtain: $=(x+5)(2x+3)$ This means that the equation above can be written as: $(x+5)(2x+3)=0$ Use the Zero-Factor Property by equating each factor to zero. Then, solve each equation to obtain: \begin{array}{ccc} &x+5=0 &\text{or} &2x+3=0 \\&x=-5 &\text{or} &2x=-3 \\&x=-5 &\text{or} & x=-\frac{3}{2} \end{array} Checking: If x=-5: $2(-5)^2+13(-5)+10=-5? \\2(25)+(-65)+10=-5? \\50-65+10=-5? \\-5=-5$ If $x=-\frac{3}{2}$: $2(-\frac{3}{2})^2+13(-\frac{3}{2})+10=-5? \\2(\frac{9}{4}-\frac{39}{2}+10=-5? \\\frac{9}{2} -\frac{39}{2} + \frac{20}{2} =-5? \\\frac{-10}{2}=-5? \\-5=-5$ Thus, the solution set is $\color{blue}{\left\{5, -\frac{3}{2}\right\}}$.