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: 19

Answer

zero: $2,$ $P(x)=(x-2)^{3}$

Work Step by Step

$P(x)=x^{3}-6x^{2}+12x-8$ ($3$ sign variations, we expect $1$ or $3$ positive real zeros) $P(-x)=-x^{3}-6x^{2}-12x-8$ ( no sign variations, we expect no negative real zeros) $a_{0}=-8 \qquad $p: $\pm 1,\pm 2,\pm 4,\pm 8,$ $a_{n}=1,\qquad $q: $\pm 1, $ Possible rational zeros $\displaystyle \frac{p}{q}:\quad \pm 1,\pm 2,\pm 4,\pm 8,$ Test $x=1 $ by synthetic division: $\left[\begin{array}{lllllll} \underline{1|} & 1 & -6 & 12 & -8 & & \\ & & 1 & -5 & -7 & & \\ & -- & -- & -- & -- & & \\ & 1 & -5 & -7 & \underline{|-15} & \Rightarrow & P(1)\neq 0, \end{array}\right.$ $1$ is not a zero, test $x=2$ $\left[\begin{array}{lllllll} \underline{2|} & 1 & -6 & 12 & -8 & & \\ & & 2 & -8 & 8 & & \\ & -- & -- & -- & -- & & \\ & 1 & -4 & 4 & \underline{|0} & \Rightarrow & P(2)=0, \end{array}\right.$ $x=2$ is a zero, $(x-2)$ is a factor of $P(x).$ The bottom row gives coefficients of the quotient. $P(x)=x^{3}-3x-4=(x-2)(x^{2} -4x+4)$ We can factor the second parentheses: recognize a square of a difference of x and 2 $P(x)=(x-2)(x-2)^{2}=(x-2)^{3}$ So, the zero is 2
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