Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises - Page 355: 28


$x=\left\{ -5,-1 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ (x+8)(x-2)=-21 ,$ in $ax^2+bx+c=0$ form. Then, express the equation in factored form. Next, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x(x)+x(-2)+8(x)+8(-2)=-21 \\\\ x^2-2x+8x-16=-21 \\\\ x^2+(-2x+8x)+(-16+21)=0 \\\\ x^2+6x+5=0 .\end{array} To factor the equation above, find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Then, express the factored form as $(x+m_1)(x+m_2)=0.$ In the equation above, the value of $c$ is $ 5 $ and the value of $b$ is $ 6 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,5 \} \\ \{ -1,-5 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 1,5 \}.$ Hence, the factored form of the equation above is \begin{array}{l}\require{cancel} (x+1)(x+5)=0 .\end{array} Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then, \begin{array}{l}\require{cancel} x+1=0 \text{ OR } x+5=0 .\end{array} Solving each of the equations above results to \begin{array}{l}\require{cancel} x+1=0 \\\\ x=-1 \\\\\text{ OR }\\\\ x+5=0 \\\\ x=-5 .\end{array} Hence, the solutions are $ x=\left\{ -5,-1 \right\} .$
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