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 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set - Page 326: 53



Work Step by Step

Factoring the negative $GCF= -1 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -9x^2-18x-5 \\\\= -(9x^2+18x+5) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} -(9x^2+18x+5) \end{array} has $ac= 9(5)=45 $ and $b= 18 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 15,3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -(9x^2+15x+3x+5) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} -[(9x^2+15x)+(3x+5)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -[3x(3x+5)+(3x+5)] .\end{array} Factoring the $GCF= (3x+5) $ of the entire expression above results to \begin{array}{l}\require{cancel} -[(3x+5)(3x+1)] \\\\= -(3x+5)(3x+1) .\end{array}
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