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: 52

Answer

$x\displaystyle \in\left\{ 1-\sqrt 3, 1+\sqrt 3, -\frac{1}{3} \right\}$

Work Step by Step

See The Rational Zero Theorem: If $\frac{p}{q}$ is a zero of the polynomial $f$ with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$. ------------------------ $f(x)=3x^{3}-5x^{2}-8x-2$ Candidates for zeros, $\frac{p}{q}:$ $p:\qquad \pm 1, \pm 2$ $q:\qquad \pm 1, \pm3$ $\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{2}{3}$ Try for $x=-\frac{1}{3}:$ $\begin{array}{lllll} \underline{-\frac{1}{3}}|& 3 & -5 & -8 & -2\\ & & -1& 2& 2\\ & -- & -- & -- & --\\ & 3 & -6& -6 & |\underline{0} \end{array}$ $-\frac{1}{3}$ is a zero, $f(x)=(x+\frac{1}{3})(3x^2-6x-6)$ Solving for the trinomial, using the quadratic formula for the quadratic equation of $ax^2+bx+c$, $x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}$. in this case, $3x^2-6x-6$, $x=\frac{6\pm \sqrt {(-6)^2-4 \times 3\times (-6)}}{2\times 3}=\frac{6\pm 6\sqrt {3}}{6}=1\pm\sqrt 3$. Therefore, the real zeros of the equations are: $x\displaystyle \in\left\{ 1-\sqrt 3, 1+\sqrt 3, -\frac{1}{3} \right\}$
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