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: 35



Work Step by Step

Factoring the $GCF= 2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 10a^2-8a-18 \\\\= 2(5a^2-4a-9) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 2(5a^2-4a-9) \end{array} has $ac= 5(-9)=-45 $ and $b= -4 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -9,5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2(5a^2-9a+5a-9) .\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} 2[(5a^2-9a)+(5a-9)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2[a(5a-9)+(5a-9)] .\end{array} Factoring the $GCF= (5a-9) $ of the entire expression above results to \begin{array}{l}\require{cancel} 2[(5a-9)(a+1)] \\\\= 2(5a-9)(a+1) .\end{array}
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