Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.1 - Complex Numbers - Exercise Set - Page 314: 11

Answer

The standard form of the expression $\left( -5+4i \right)\left( 3+i \right)$ is $-19+7i$.

Work Step by Step

Consider the expression $\left( -5+4i \right)\left( 3+i \right)$. Use the FOIL method. $\left( -5+4i \right)\left( 3+i \right)=-15-5i+12i+4{{i}^{2}}$ Replace the value ${{i}^{2}}=-1$. $\left( -5+4i \right)\left( 3+i \right)=-15-5i+12i+4\left( -1 \right)$ Make a group of real and imaginary terms. $\left( -5+4i \right)\left( 3+i \right)=-15-4+12i-5i$ Simplify the real and imaginary terms. $\begin{align} & \left( -5+4i \right)\left( 3+i \right)=\left( -15-4 \right)+\left( 12-5 \right)i \\ & =-19+7i \end{align}$ Therefore, the standard form of the expression $\left( -5+4i \right)\left( 3+i \right)$ is $-19+7i$.
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