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 320: 24

Answer

$P(x)=(x-2)(x+3)(x-5)$ Zeros: $ \ -3,\ 2, \ 5 \ $

Work Step by Step

$P(x)=x^{3}-4x^{2}-11x+30$ $P(-x)=-x^{3}-4x^{2}+11x+30$ Descart's rule of signs: P(x) has 2 sign changes $\Rightarrow$ 2 or 0 positive zeros. P(-x) has 1 sign changes $\Rightarrow$ 1 negative zeros. Rational Zeros Theorem: candidates for p:$\quad \pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30$ candidates for q:$\quad \pm 1,$ Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30$ Testing with synthetic division, $\left.\begin{array}{l} 2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr} 1 & -4 & -11 &30\\\hline & 2 & 4 & -30\\\hline 1& 2 & -15&|\ \ 0\end{array}$ $P(x)=(x-2)(x^{2}+2x-15)$ For the trinomial, find two factors of $-15$ with sum $2$ ... ( they are $+3$ and $-5)$ $P(x)=(x-2)(x+3)(x-5)$ Zeros: $ \ -3,\ 2, \ 5 \ $
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