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

Published by Pearson

# Chapter 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set - Page 686: 82

#### Answer

$\dfrac{1-12i}{15}$

#### Work Step by Step

Multiplying by the conjugate of the denominator, the given expression, $\dfrac{5-2i}{3+6i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{5-2i}{3+6i}\cdot\dfrac{3-6i}{3-6i} \\\\= \dfrac{5(3)+5(-6i)-2i(3)-2i(-6i)}{(3)^2-(6i)^2} \\\\= \dfrac{15-30i-6i+12i^2}{9-36i^2} \\\\= \dfrac{15-30i-6i+12(-1)}{9-36(-1)} \\\\= \dfrac{15-30i-6i-12}{9+36} \\\\= \dfrac{(15-12)+(-30i-6i)}{9+36} \\\\= \dfrac{3-36i}{45} \\\\= \dfrac{3(1-12i)}{45} \\\\= \dfrac{\cancel{3}(1-12i)}{\cancel{3}\cdot15} \\\\= \dfrac{1-12i}{15} \end{array}

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