## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.7 - Complex Numbers - 7.7 Exercises - Page 491: 64

#### Answer

$25$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(2-i)^2(2+i)^2 ,$ use the laws of exponents to express the factors in a single exponent. Then use the special product on multiplying the sum and difference of like terms. Also, use $i^2=-1.$ $\bf{\text{Solution Details:}}$ Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} [(2-i)(2+i)]^2 .\end{array} Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} [(2)^2-(i)^2]^2 \\\\= [4-i^2]^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} [4-(-1)]^2 \\\\= [4+1]^2 \\\\= ^2 \\\\= 25 .\end{array}

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