College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 3, Polynomial and Rational Functions - Section 3.4 - Real Zeros of Polynomials - 3.4 Exercises - Page 319: 16

Answer

$P(x)=(x+1)(x+2)(x-7)$ Zeros: $-2,\ -1, \ 7 \ \ \ $

Work Step by Step

$P(x)=x^{3}-4x^{2}-19x-14$ $P(-x)=-x^{3}-4x^{2}+19x-14$ Decscart's rule of signs: P(x) has $1$ sign changes $\Rightarrow 1$ positive zeros. P(-x) has $2$ sign changes $\Rightarrow$ 2 or 0 negative zeros. Rational Zeros Theorem: candidates for p:$\quad \pm 1,\pm 2,\pm 7,\pm 14$ candidates for q:$\quad \pm 1,$ Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 7,\pm 14$ Testing with synthetic division, $\left.\begin{array}{l} -1|\\ \\ \\ \end{array}\right.\begin{array}{rrr} -1 &-4&-19&-14\\\hline &-1& 5 &14\\\hline 1& -5&-14&|\ \ 0\end{array}$ Remainder th: $x=-1$ is a zero $\Rightarrow (x+1)$ is a factor (Factor th.). $P(x)=(x+1)(x^{2}-5x-14)$ For the trinomial, find two factors of $-14$ with sum $-5$ ... ( they are $+2$ and $-7)$ $P(x)=(x+1)(x+2)(x-7)$ Zeros: $-2\ -1, \ 7 \ \ \ $
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