Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 3 - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 293: 10

Answer

$(a)\quad $Zeros: $0$ and $-\displaystyle \frac{1}{2}\pm\frac{\sqrt{3}}{2}i$ $(b)\quad P(x)=x(x +\displaystyle \frac{1}{2}-\frac{\sqrt{3}}{2}i)(x +\displaystyle \frac{1}{2}+\frac{\sqrt{3}}{2}i)$

Work Step by Step

Complete Factorization Theorem (p.287) $P$ factors into $n$ linear factors : $P(x)=a(x-c_{1})(x-c_{2})\cdots(x-c_{n})$ where $a$ is the leading coefficient of $P$ and $c_{1}, c_{1}, \ldots, c_{n}$ are the zeros of $P$. ----- $(a)\quad$ $x^{3}+x^{2}+x=0$ $x(x^{2}+x+1)=0$ So, either $x=0\Rightarrow\quad x=0$ is a zero, or $x^{2}+x+2=0\displaystyle \Rightarrow\quad x=\frac{-1\pm\sqrt{1-4}}{2}=\frac{-1\pm i\sqrt{3}}{2}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$ Zeros: $0$ and $-\displaystyle \frac{1}{2}\pm\frac{\sqrt{3}}{2}i$ $(b)$ The leading coefficient of P is $a=1$, so, by the above theorem, $P(x)=x[x-( -\displaystyle \frac{1}{2}+\frac{\sqrt{3}}{2}i)][x -(-\displaystyle \frac{1}{2}-\frac{\sqrt{3}}{2}i)]$ $P(x)=x(x +\displaystyle \frac{1}{2}-\frac{\sqrt{3}}{2}i)(x +\displaystyle \frac{1}{2}+\frac{\sqrt{3}}{2}i)$
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