Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.3 Complex Numbers - 1.3 Exercises: 71

Answer

$(2+i)(2-i)(4+3i)=20+15i$

Work Step by Step

$(2+i)(2-i)(4+3i)$ Begin by evaluating the product between the first two factors. The product of the first two factors represents the factored form of a difference of two squares. $(2+i)(2-i)(4+3i)=(4-i^{2})(4+3i)=...$ Substitute $i^{2}$ by $-1$ and simplify the first factor: $...=[4-(-1)](4+3i)=(4+1)(4+3i)=5(4+3i)=...$ Evaluate the remaining product: $...=20+15i$
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