## College Algebra (11th Edition)

Published by Pearson

# Chapter 1 - Review Exercises: 29

#### Answer

$1-2i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\dfrac{-12-i}{-2-5i} ,$ multiply the numerator and the denominator by the conjugate of the denomniator. Use special products to multiply the resulting expression. Then use the equivalence $i^2=-1.$ Finally, combine the real parts and the imaginary parts. $\bf{\text{Solution Details:}}$ Reversing the operator between the terms of the expression, $-2-5i ,$ then its conjugate is $-2+5i .$ Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{-12-i}{-2-5i}\cdot\dfrac{-2+5i}{-2+5i} \\\\= \dfrac{(-12-i)(-2+5i)}{(-2-5i)(-2+5i)} .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{-12(-2)-12(5i)-i(-2)-i(5i)}{(-2-5i)(-2+5i)} \\\\= \dfrac{24-60i+2i-5i^2}{(-2-5i)(-2+5i)} .\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} \dfrac{24-60i+2i-5i^2}{(-2)^2-(5i)^2} \\\\= \dfrac{24-60i+2i-5i^2}{4-25i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{24-60i+2i-5(-1)}{4-25(-1)} \\\\= \dfrac{24-60i+2i+5}{4+25} .\end{array} Combining the real parts and the imaginary parts results to \begin{array}{l}\require{cancel} \dfrac{(24+5)+(-60i+2i)}{4+25} \\\\= \dfrac{29-58i}{29} \\\\= \dfrac{\cancel{29}-\cancel{29}(2)i}{\cancel{29}} \\\\= 1-2i .\end{array}

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