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: 33

Answer

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

Work Step by Step

Factoring the $GCF= -3 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -6x^2-33x-15 \\\\= -3(2x^2+11x+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+11x+5) \end{array} has $ac= 2(5)=10 $ and $b= 11 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 10,1 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -3(2x^2+10x+1x+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+10x)+(1x+5)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -3[2x(x+5)+(x+5)] .\end{array} Factoring the $GCF= (x+5) $ of the entire expression above results to \begin{array}{l}\require{cancel} -3[(x+5)(2x+1)] \\\\= -3(x+5)(2x+1) .\end{array}
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