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: 49

Answer

$x\displaystyle \in\left\{ \frac{1- \sqrt {5}}{2}, \frac{1+ \sqrt {5}}{2},3\right\}$

Work Step by Step

See The Rational Zero Theorem: If $\frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$. ------------------------ $f(x)=x^4-7x^{3}+14x^{2}-3x-9$ Candidates for zeros, $\frac{p}{q}:$ $p:\qquad \pm 1, \pm 3, \pm9$ $q:\qquad \pm 1$ $\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 3, \pm 9$ Try for $x=3:$ $\begin{array}{lllll} \underline{3}|& 1 & -7 & 14 & -3 & -9\\ & & 3 & -12& 6& 9\\ & -- & -- & -- & --\\ & 1 & -4 & 2& 3 & |\underline{0} \end{array}$ $3$ is a zero, $f(x)=(x-3)(x^3-4x^2+2x+3)$ Try for $x=3$: $\begin{array}{lllll} \underline{3}|& 1 & -4 & 2 & 3\\ & & 3& -3& -3\\ & -- & -- & -- & --\\ & 1 & -1& -1 & |\underline{0} \end{array}$ $3$ is a zero, $f(x)=(x-3)^2(x^2-x-1)$ 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, $x^2+x-1$, $x=\frac{1\pm \sqrt {-1^2-4 \times 1\times (-1)}}{2\times 1}=\frac{1\pm \sqrt {5}}{2}$. Therefore, the real zeros of the equations are: $x\displaystyle \in\left\{ \frac{1- \sqrt {5}}{2}, \frac{1+ \sqrt {5}}{2},3\right\}$
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