Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.5 Solving Equations by Factoring - 4.5 Exercises - Page 360: 42


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

Work Step by Step

Write the equation in the form $ax^2+bx+c=0$ to obtain: $$x^2+11x+28=0$$ Recall: If the trinomial $x^2+bx+c$ is factoable, then its factored form is $(x+d)(x+e)$ where $c=de$ and $b=d+e$. Look for factors of $28$ whose sum is equal to $11$. Note that $28=7(4)$ and $11=7+4$. Thus, $d=7$ and $e=4$. Hence, $$x^2+11x+28=(x+7)(x+4)$$ The given equation is equivalent to $(x+7)(x+4)=0$. Use the Zero-Product Property by equating each factor to zero, then solving each equation to obtain: \begin{align*} x+7&=0 &\text{or}& &x+4=0\\ x&=-7 &\text{or}& &x=-4\\ \end{align*}
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