Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 3 - Section 3.4 - Real Zeros of Polynomials - 3.4 Exercises - Page 283: 18

Answer

zeros: $-1 $ and $2$ $P(x)=(x-2)(x+1)^{2}$

Work Step by Step

$P(x)=x^{3}-3x-2$ (1 sign variation, we expect 1 positive real zero) $a_{0}=-2 \qquad $p: $\pm 1,\pm 2,$ $a_{n}=1,\qquad $q: $\pm 1, $ Possible rational zeros $\displaystyle \frac{p}{q}:\quad \pm 1,\pm 2$. Test $x=1 $ by synthetic division (with 0 as the missing coefficient) : $\left[\begin{array}{lllllll} \underline{1|} & 1 & 0 & -3 & -2 & & \\ & & 1 & 1 & -2 & & \\ & -- & -- & -- & -- & & \\ & 1 & 1 & -2 & \underline{|-4} & \Rightarrow & P(1)=-4, \end{array}\right.$ $1$ is not a zero, test $x=2$ $\left[\begin{array}{lllllll} \underline{2|} & 1 & 0 & -3 & -2 & & \\ & & 2 & 4 & 2 & & \\ & -- & -- & -- & -- & & \\ & 1 & 2 & 1 & \underline{|0} & \Rightarrow & P(2)=0, \end{array}\right.$ $x=2$ is a zero, $(x-2)$ is a factor of $P(x).$ The bottom row gives coefficients of the quotient. $P(x)=x^{3}-3x-4=(x-2)(x^{2}+2x+1)$ We can factor the second parentheses: recognize a square of a sum of x and $1$ $P(x)=(x-2)(x+1)^{2}$ So, the zeros are $-1 $ and $2$
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