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: 51

Answer

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

Work Step by Step

$P(x)=4x^{3}-6x^{2}+1,$ ($2$ sign variations, we expect $3$ or $1$ positive real zero) $P(-x)=-4x^{3}-6x^{2}+1$ ($1$ sign variations, we expect $1$ negative real zeros) $a_{0}=1 \quad $p: $\pm 1,$ $a_{n}=4,\qquad $q: $\pm 1,\pm 2,\pm 4$ Possible rational zeros $\displaystyle \frac{p}{q}:\quad$ with synthetic division, we try$ \pm$1, $\pm 1/2$... $\begin{array}{r|rrrrrrrrrrr}\hline 1/2 & 4 & -6 & 0 & 1 & & & \\ & & 2 & -2 & -1 & & & \\ \hline & 4 & -4 & -2 & 0 & & & \\ \end{array}$ $P(x)=(x+\displaystyle \frac{1}{2})(4x^{2}-4x-2)$ Find the zeros of the second factor with the quadratic formula $x=\displaystyle \frac{4\pm\sqrt{(-4)^{2}-4(4)(-2)}}{2(4)}=\frac{4\pm\sqrt{48}}{8}=\frac{4\pm 4\sqrt{3}}{8}=\frac{1\pm\sqrt{3}}{2}$ zeros:$\displaystyle \qquad \frac{1}{2},\ \ \ \frac{1\pm\sqrt{3}}{2}$
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