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: 38

Answer

$3(2x+5)(x+1)$

Work Step by Step

Factoring the $GCF= 3 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 6x^2+21x+15 \\\\= 3(2x^2+7x+5) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 3(2x^2+7x+5) \end{array} has $ac= 2(5)=10 $ and $b= 7 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 5,2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3(2x^2+5x+2x+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} 3[(2x^2+5x)+(2x+5)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3[x(2x+5)+(2x+5)] .\end{array} Factoring the $GCF= (x+6) $ of the entire expression above results to \begin{array}{l}\require{cancel} 3[(2x+5)(x+1)] \\\\= 3(2x+5)(x+1) .\end{array}
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