Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 41


$a=\left\{ -2,6 \right\}$

Work Step by Step

Expressing the given equation in the form $ax^2+bx+c=0,$ the given equation is equivalent to \begin{array}{l}\require{cancel} (a+1)(a-5)=7 \\\\ a(a)+a(-5)+1(a)+1(-5)=7 \\\\ a^2-5a+a-5=7 \\\\ a^2-5a+a-5-7=0 \\\\ a^2-4a-12=0 .\end{array} Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{ equation }$ \begin{array}{l}\require{cancel} a^2-4a-12=0 \end{array} has $c= -12 $ and $b= -4 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -6,2 \right\}.$ Using these two numbers, the $\text{ equation }$ above is equivalent to \begin{array}{l}\require{cancel} (a-6)(a+2)=0 .\end{array} Equating each factor to zero (Zero Product Property), and then isolating the variable, the solutions are $ a=\left\{ -2,6 \right\} .$
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