Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 3 - Polynomial and Rational Functions - Section 3.3 Complex Zeros; Fundamental Theorem of Algebra - 3.3 Assess Your Understanding - Page 231: 31

Answer

$x=1,\frac{-1\pm i\sqrt {3}}{2}$ $f(x)=(x-1)(x+\frac{1+ i\sqrt {3}}{2})(x+\frac{1- i\sqrt {3}}{2})$

Work Step by Step

Step 1. $x^3-1=0 \Longrightarrow (x-1)(x^2+x+1)=0 \Longrightarrow x=1\ and\ x=\frac{-1\pm\sqrt {1-4}}{2}=\frac{-1\pm i\sqrt {3}}{2}$ Step 2. We have $f(x)=(x-1)(x+\frac{1+ i\sqrt {3}}{2})(x+\frac{1- i\sqrt {3}}{2})$
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