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: 17

Answer

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

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. $k=2$ $\left\{\begin{array}{rrrrrrr} 2 & | & 2 & -3 & -17 & 30 & \\ & & & 4 & 2 & -30 & \\ & & -- & -- & -- & -- & --\\ & & 2 & 1 & -15 & \fbox{$0$} & \end{array}\right.$ $f(x)= (x-2)(2x^{2}+x-15)$ 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 $2(-15)=-30$ whose sum is $1$ ... are $+6$ and $-5$: $2x^{2}+x-15\\ =2x^{2}+6x-5x-15$ $=2x(x+3)-5(x+3)\\=(x+3)(2x-5)$ So, $f(x)= (x-2)(x+3)(2x-5)$
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