Answer
$f(x)=\dfrac{1}{2}(x+1)(x-2)(x-3)$
Work Step by Step
Here, we have $f(x)=a (x+1)(x-2)(x-3)$
When $x=0$, then we have $3=a (0+1)(0-2)(0-3)$
or, $6a=3$
This gives: $a=\dfrac{1}{2}$
Thus, we have $f(x)=\dfrac{1}{2}(x+1)(x-2)(x-3)$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 5 Polynomials and Polynomial Functions - 5.9 Write Polynomial Functions and Models - 5.9 Exercises - Skill Practice - Page 397: 3
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