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: 49

Answer

zeros:$\displaystyle \qquad 3,\ \frac{1\pm\sqrt{5}}{2}$

Work Step by Step

$P(x)=x^{4}-7x^{3}+14x^{2}-3x-9,$ ($3$ sign variations, we expect $3$ or $1$ positive real zero) $P(-x)=x^{4}+7x^{3}+14x^{2}+3x-9,$ (1 sign variations, we expect 1 negative real zeros) $a_{0}=-9 \quad $p: $\pm 1,\pm 3,\pm 9$ $a_{n}=1,\qquad $q: $\pm 1,$ Possible rational zeros $\displaystyle \frac{p}{q}:\quad$ with synthetic division, we try$ \pm$1, $\pm 3$... $\begin{array}{r|rrrrrrrrrrr}\hline 3 & 1 & -7 & 14 & -3 & -9 & & \\ & & 3 & -12 & 6 & 9 & & \\ \hline & 1 & -4 & 2 & 3 & 0 & & \\ & & & & & & & \\ 3 & 1 & -4 & 2 & 3 & & & \\ & & 3 & -3 & -3 & & & \\ \hline & 1 & -1 & -1 & 0 & & & \\ \end{array}$ $P(x)=(x-3)^{2}(x^{2}-x-1)$ 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:$\displaystyle \qquad 3,\ \frac{1\pm\sqrt{5}}{2}$
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