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: 22

Answer

$f(x)= (x+5)(2x+3)(4x-1)$

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. $k=-5$ $\left\{\begin{array}{rrrrrrr} -5 & | & 8 & 50 & 47 & -15 & \\ & & & -40 & -50 & 15 & \\ & & -- & -- & -- & -- & --\\ & & 8 & 10 & -3 & \fbox{$0$} & \end{array}\right.$ $f(x)= (x+5)(8x^{2}+10x-3)$ We now factor the trinomial $ax^{2}+bx+c$ by searching for two factors of $ac$ whose sum is $b.$ If they exist, rewrite $bx$ and factor in pairs. Factors of $8(-3)=-24$ whose sum is $+10$ ... are $+12$ and $-2$: $8x^{2}+10x-3\\=8x^{2}+12x-2x-3$ $=4x(2x+3)-(2x+3)\\=(2x+3)(4x-1)$ So, $f(x)= (x+5)(2x+3)(4x-1)$
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