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: 43


$x=\left\{ -11,5 \right\}$

Work Step by Step

Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{ equation }$ \begin{array}{l}\require{cancel} x^2+6x-55 \end{array} has $c= -55 $ and $b= 6 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 11,-5 \right\}.$ Using these two numbers, the $\text{ equation }$ above is equivalent to \begin{array}{l}\require{cancel} (x+11)(x-5)=0 .\end{array} Equating each factor to zero (Zero Product Property), and then isolating the variable, the solutions are $ x=\left\{ -11,5 \right\} .$
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