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: 29

Answer

The zeros are $3$, (which is given,) $-1+i$ and $-1-i.$

Work Step by Step

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. $k=3$ $\left\{\begin{array}{lllllll} 3 & | & 1 & -1 & -4 & -6 & \\ & & & 3 & 6 & 6 & \\ & & -- & -- & -- & -- & --\\ & & 1 & 2 & 2 & \fbox{$0$} & \end{array}\right.$ $f(x)= (x-3)(x^{2}+2x+2)$ We can use the quadratic formula to find the zeros of the quotient. $x=\displaystyle \frac{b\pm\sqrt{b^{2}-4ac}}{2a} =\frac{-2\pm\sqrt{4-4(1)(2)}}{2(1)}$ $=\displaystyle \frac{-2\pm\sqrt{-4}}{2}=\frac{-2\pm 2i}{2}=\frac{2(-1\pm i)}{2}$ $=-1\pm i$ The zeros are $3$, (which is given,) $-1+i$ and $-1-i.$
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