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 388: 63

Answer

$x\in\left\{-2,-3\right\}$

Work Step by Step

We try to find rational zeros of $f(x)=x^{4}+4x^{3}+2x^{2}-x+6$ Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1}$ Testing with synthetic division, ... we try $x+2$ $\left.\begin{array}{l} -2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 1&4 & 2 & -1 &6 \\\hline &-2 &-4 & 4 & -6 \\\hline 1&2 & -2 &3 & |\ \ 0 \end{array}$ $f(x)=(x+2)(x^{3}+2x^{2}-2x+3)$ Testing with synthetic division, ... we try $x+3$ $\left.\begin{array}{l} -3 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 1 & 2 & -2 &3 \\\hline &-3 & 3 & -3 \\\hline 1& -1 & 1 & |\ \ 0 \end{array}$ $f(x)=(x+2)(x+3)(x^{2}-x+1)$ The discriminant, $b^{2}-4ac=1-4=-3$ for the equation $ x^{2}-x+1=0\quad$ is negative, so no further real zeros, as the equation does not have real solutions. $f(x)=(x+2)(x+3)(x^{2}-x+1)=0$ when $x\in\left\{-2,-3\right\}$
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