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 284: 26

Answer

$P(x)=(x-1)^{2}(x+2)(x-2)$ Zeros: $\pm 2, 1$

Work Step by Step

$P(x)=x^{4}-2x^{3}-3x^{2}+8x-4,$ ($3$ sign variation, we expect $1$ or $3$ positive real zero) $P(-x)=x^{4}+2x^{3}-3x^{2}-8x-4,$ ($1$ sign variations, we expect $1$ negative real zeros) $a_{0}=-4, \quad $p: $\pm 1,\pm 2,\pm 4,$ $a_{n}=1,\qquad $q: $\pm 1. $ Possible rational zeros $\displaystyle \frac{p}{q}:\quad$ $\pm 1,\pm 2,\pm 4.$ Test $x= 1 $ by synthetic division: $\left[\begin{array}{lllllll} \underline{1|} & 1 & -2 & -3 & 8 & -4 & \\ & & 1 & -1 & -4 & 4 & \\ & -- & -- & -- & -- & -- & \\ & 1 & -1 & -4 & 4 & \underline{|0} & \Rightarrow P(1)=0, \end{array}\right.$ $P(x)=(x-1)(x^{3}-x^{2}-4x+4)$ To factor the parentheses, group terms two by two, $x^{3}-x^{2}-4x+4=x^{2}(x-1)-4(x-1)$ $=(x-1)(x^{2}-4)$ ... difference of squares ... $=(x-1)(x+2)(x-2)$ $P(x)=(x-1)(x-1)(x+2)(x-2)$ $P(x)=(x-1)^{2}(x+2)(x-2)$ Zeros: $\pm 2, 1$
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