## 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: 67

#### Answer

$(a-2b)(2a-b)$

#### Work Step by Step

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

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