## 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: 64

#### Answer

$-2a^2(5a-2)(7a-4)$

#### Work Step by Step

Factoring the negative $GCF= -2a^2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -70a^4+68a^3-16a^2 \\\\= -2a^2(35a^2-34a+8) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} -2a^2(35a^2-34a+8) \end{array} has $ac= 35(8)=280$ and $b= -34 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -14,-20 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -2a^2(35a^2-14a-20a+8) .\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} -2a^2[(35a^2-14a)-(20a-8)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -2a^2[7a(5a-2)-4(5a-2)] .\end{array} Factoring the $GCF= (5a-2)$ of the entire expression above results to \begin{array}{l}\require{cancel} -2a^2[(5a-2)(7a-4)] \\\\= -2a^2(5a-2)(7a-4) .\end{array}

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