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

Published by Pearson

# Chapter 4 - Polynomials - 4.7 Polynomials in Several Variables - 4.7 Exercise Set - Page 284: 67

#### Answer

$a^2+2ab+b^2-c^2$

#### Work Step by Step

Grouping the first 2 terms of each trinomial factor, the given expression, $(a+b-c)(a+b+c) ,$ is equivalent to \begin{array}{l}\require{cancel} [(a+b)-c][(a+b)+c] .\end{array} Using $(a+b)(a-b)=a^2-b^2$ or the product of the sum and difference of like terms, the expression above simplifies to \begin{array}{l}\require{cancel} (a+b)^2-(c)^2 \\\\= (a+b)^2-c^2 .\end{array} Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the expression above simplifies to \begin{array}{l}\require{cancel} (a)^2+2(a)(b)+(b)^2-c^2 \\\\= a^2+2ab+b^2-c^2 .\end{array}

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