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: 15

Answer

$P(x)=(x-1)(x+5)(x-2)$ Zeros: $\ \ -5,\ 1, \ 2 \ \ \ $

Work Step by Step

$P(x)=x^{3}+2x^{2}-13x+10$ $P(-x)=-x^{3}+2x^{2}+13x+10$ Decscart's rule of signs: P(x) has 2 sign changes $\Rightarrow$ 2 or 0 positive zeros. P(-x) has 1 sign change $\Rightarrow$ 1 negative zero. Rational Zeros Theorem: candidates for p:$\quad \pm 1,\pm 2,\pm 5,\pm 10$ candidates for q:$\quad \pm 1,$ Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 5,\pm 10$ Testing with synthetic division, $\left.\begin{array}{l} 1|\\ \\ \\ \end{array}\right.\begin{array}{rrr} 1 &2&-13&10\\\hline &1&3 &10\\\hline 1&3&-10&|\ \ 0\end{array}$ Remainder th: $x=1$ is a zero $\Rightarrow (x-1)$ is a factor (Factor th.). $P(x)=(x-1)(x^{2}+3x-10)$ For the trinomial, find two factors of $-10$ with sum $3$ ... ( they are $+5$ and $-2)$ $P(x)=(x-1)(x+5)(x-2)$ Zeros: $-5,\ 1, \ 2 \ \ \ $
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