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: 46

Answer

$x\displaystyle \in\left\{-1, \frac{-1- \sqrt {61}}{6}, \frac{-1+ \sqrt {61}}{6}, 3\right\}$

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, $a_{n}$. ------------------------ $f(x)=3x^4-5x^{3}-16x^{2}+7x+15$ Candidates for zeros, $\frac{p}{q}:$ $p:\qquad \pm 1, \pm 3, \pm 5, \pm15$ $q:\qquad \pm 1, \pm 3$ $\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 3, \pm5, \pm15 ,\pm\frac{1}{3}, \pm \frac{5}{3}$ Try for $x=-1:$ $\begin{array}{lllll} \underline{-1}|& 3 & -5 & -16 & 7 & 15\\ & & -3 & 8& 8& -15\\ & -- & -- & -- & --\\ & 3 & -8 & -8& 15 & |\underline{0} \end{array}$ $-1$ is a zero, $f(x)=(x+1)(3x^3-8x^2-8x+15)$ Try for $x=3$: $\begin{array}{lllll} \underline{3}|& 3 & -8 & -8 & 15\\ & & 9& 3& -15\\ & -- & -- & -- & --\\ & 3 & 1& -5 & |\underline{0} \end{array}$ $3$ is a zero, $f(x)=(x+1)(x-3)(3x^2+x-5)$ 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, $3x^2+x-5$, $x=\frac{-1\pm \sqrt {1^2-4 \times 3\times (-5)}}{2\times 3}=\frac{-1\pm \sqrt {61}}{6}$. Therefore, the real zeros of the equation are: $x\displaystyle \in\left\{-1, \frac{-1- \sqrt {61}}{6}, \frac{-1+ \sqrt {61}}{6}, 3\right\}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.