## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set - Page 326: 72

#### Answer

$-3(2a+5b)(5a+2b)$

#### Work Step by Step

Factoring the negative $GCF= -3 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -30a^2-87ab-30b^2 \\\\= -3(10a^2+29ab+10b^2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} -3(10a^2+29ab+10b^2) \end{array} has $ac= 10(10)=100$ and $b= 29 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 25,4 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -3(10a^2+25ab+4ab+10b^2) .\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[(10a^2+25ab)+(4ab+10b^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -3[5a(2a+5b)+2b(2a+5b)] .\end{array} Factoring the $GCF= (3x-4y)$ of the entire expression above results to \begin{array}{l}\require{cancel} -3[(2a+5b)(5a+2b)] \\\\= -3(2a+5b)(5a+2b) .\end{array}

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