## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises: 30

#### Answer

$(2+k-h)(2-k+h)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $-k^2-h^2+2kh+4 ,$ rewrite and group terms that will form a perfect square trinomial. Then factor the trinomial. The resulting expression becomes a difference of $2$ squares. Factor this expression using the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The given expression can be rewritten as \begin{array}{l}\require{cancel} 4-k^2+2kh-h^2 .\end{array} Grouping the last $3$ terms above results to \begin{array}{l}\require{cancel} 4-(k^2-2kh+h^2) .\end{array} The trinomial above is a perfect square trinomial. Using $a^2\pm2ab+b^2=(a\pm b)^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 4-(k-h)^2 .\end{array} The expression above is a difference of $2$ squares. Using $a^2-b^2=(a+b)(a-b),$ the factored form of the expression above is \begin{array}{l}\require{cancel} [2+(k-h)][2-(k-h)] \\\\= [2+k-h][2-k+h] \\\\= (2+k-h)(2-k+h) .\end{array}

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