# Chapter 4 - Quadratic Functions - 4.5 Solving Equations by Factoring - 4.5 Exercises - Page 359: 20

$\color{blue}{\left\{-7, 4\right\}}$

#### Work Step by Step

RECALL: A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$. The trinomial's factored form will be: $x^2+bx+c=(x+d)(x+e)$ The trinomial in the given equation has $b=3$ and $c=-28$. Note that $-28=7(-4)$ and $3= 7+(-4)$. This means that $d=7$ and $e=-4$ Thus, the factored form of the trinomial is: $(p+7)[p+(-4)]=(p+7)(p-4)$ The given equation maybe written as: $(p+7)(p-4)=0$ Use the Zero-Factor Property by equating each factor to zero. Then, solve each equation to obtain: \begin{array}{ccc} &p+7 = 0 &\text{ or } &p-4=0 \\&p=-7 &\text{ or } &p=4 \end{array} Thus, the solution set is $\color{blue}{\left\{-7, 4\right\}}$.

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