Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Review - Exercises: 80

Answer

The solutions are $x=2$, $x=2i$ and $x=-2i$

Work Step by Step

$x^{3}-2x^{2}+4x-8=0$ Group the first two terms and the last two terms together: $(x^{3}-2x^{2})+(4x-8)=0$ Take out common factor $x^{2}$ from the first parentheses and common factor $4$ from the second parentheses: $x^{2}(x-2)+4(x-2)=0$ Take out common factor $x-2$ from the left side: $(x-2)(x^{2}+4)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-2=0$ $x=2$ $x^{2}+4=0$ $x^{2}=-4$ $\sqrt{x^{2}}=\sqrt{-4}$ $x=\pm2i$ The solutions are $x=2$, $x=2i$ and $x=-2i$
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