College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 5 - Section 5.5 - The Real Zeros of a Polynomial Function - 5.5 Assess Your Understanding - Page 387: 51

Answer

$f(x)=(x+1)(2x-1)(x+\sqrt{3})(x-\sqrt{3})$ Zeros: $-\displaystyle \sqrt{3},\ \ -1,\ \ \frac{1}{2},\ \ \sqrt{3},$ all with multiplicity 1.

Work Step by Step

$f(x)$ has at most $4$ real zeros, as the degree is $4$. Rational zeros: $\displaystyle \frac{p}{q},$ in lowest terms, $p$ must be a factor of $ 3,\qquad \pm 1,\pm 3,$ and $q$ must be a factor of $ 2,\qquad \pm 1,\pm 2$ $\displaystyle \frac{p}{q}=\pm 1,\pm\frac{1}{2},\pm\frac{3}{2},\pm 3$ Testing with synthetic division, try $x+1$ $\left.\begin{array}{l} -1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 2 &1 &-7 &-3 &3 \\\hline &-2 &1 &6 & -3 \\\hline 2&-1 &-6 &3& |\ \ 0 \end{array}$ $f(x)=(x+1)(2x^{3}-x^{2}-6x+3)$ Try synthetic division again, $ x-\displaystyle \frac{1}{2}$ $\left.\begin{array}{l} 1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 2 &-1 &-6 &3 \\\hline &1 &0 & -3 \\\hline 2 & 0 &-6 & |\ \ 0 \end{array}$ $f(x)=(x+1)(x-\displaystyle \frac{1}{2})(2x^{2}-6)=(x+1)(x-\frac{1}{2})\cdot 2(x^{2}-3)$ $=(x+1)(2x-1)(x^{2}-3)$ Solving $x^{2}-3=0,$ $x=\pm\sqrt{3}$, we have the other two zeros. $f(x)=(x+1)(2x-1)(x+\sqrt{3})(x-\sqrt{3})$ Zeros: $-\displaystyle \sqrt{3},\ \ -1,\ \ \frac{1}{2},\ \ \sqrt{3},$ all with multiplicity 1.
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