College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 3 - Section 3.3 - Zeros of Polynomial Functions - 3.3 Exercises - Page 315: 30

Answer

The zeros are $1$, (which is given,) $\displaystyle \frac{-5-\sqrt{5}}{2}$ and $\displaystyle \frac{-5+\sqrt{5}}{2}$

Work Step by Step

x_{y} If k is a zero of f, then, by the factor theorem, (x-k) is a factor of f(x). Synthetic division gives us a quotient q(x), with degree one less than f(x). and we know the remainder is zero, so $f(x)=(x-k)\cdot q(x).$ We use this first. $k=1$, the x term is missing, $\left\{\begin{array}{lllllll} 1 & | & 1 & 4 & 0 & -5 & \\ & & & 1 & 5 & 5 & \\ & & -- & -- & -- & -- & --\\ & & 1 & 5 & 5 & \fbox{$0$} & \end{array}\right.$ $f(x)= (x-1)(x^{2}+5x+5)$ We can use the quadratic formula to find the zeros of the quotient. $x=\displaystyle \frac{b\pm\sqrt{b^{2}-4ac}}{2a} =\frac{-5\pm\sqrt{25-4(1)(5)}}{2(1)}$ $=\displaystyle \frac{-5\pm\sqrt{5}}{2}$ The zeros are $1$, (which is given,) $\displaystyle \frac{-5-\sqrt{5}}{2}$ and $\displaystyle \frac{-5+\sqrt{5}}{2}$
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