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: 48

Answer

zeros:$\displaystyle \qquad -2,\ 1,\ \frac{-1\pm\sqrt{5}}{2}$

Work Step by Step

$P(x)=x^{4}+2x^{3}-2x^{2}-3x+2,$ ($2$ sign variations, we expect $2$ or $0$ real zero) $P(-x)=x^{4}-2x^{3}-2x^{2}+3x+2,$ ($2$ sign variations, we expect $2$ or $0$ negative real zeros) $a_{0}=2 \quad $p: $\pm 1,\pm 2$ $a_{n}=1,\qquad $q: $\pm 1,$ Possible rational zeros $\displaystyle \frac{p}{q}:\quad$ with synthetic division, we try$ \pm$1, $\pm$2,... $\begin{array}{r|rrrrrrrrrrr}\hline 1 & 1 & 2 & -2 & -3 & 2 & & \\ & & 1 & 3 & 1 & -2 & & \\ \hline & 1 & 3 & 1 & -2 & 0 & & \\ & & & & & & & \\ -2 & 1 & 3 & 1 & -2 & & & \\ & & -2 & -2 & 2 & & & \\ \hline & 1 & 1 & -1 & 0 & & & \\ \end{array}$ $P(x)=(x-1)(x+2)(x^{2}+x-1)$ Find the zeros of the second factor with the quadratic formula $x=\displaystyle \frac{-1\pm\sqrt{(-1)^{2}-4(1)(-1)}}{2(1)}=\frac{-1\pm\sqrt{5}}{2}$ zeros:$\displaystyle \qquad 1,\ -2,\ \frac{-1\pm\sqrt{5}}{2}$
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