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 316: 54

Answer

$f(x)=2x^{3}-20x^{2}+64x-64$

Work Step by Step

If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem) So, $(x-2)$ is a factor of f, $(x-4)$ is a factor of f. The number of times $(x-k)$ occurs as a factor is referred to as the multiplicity of the zero.So, $(x-2)$ is a factor of f, degree=1 $(x-4)^{2}$ is a factor of f, degree = 2 (the sum of degrees is 3) Since f has degree 3, $\quad f(x)=a(x-2)(x-4)^{2}$ To find $a,$ use the given information: $f(1)=-18$ $a(1-2)(1-4)^{2}=-18$ $a(-1)(9)=-18$ $-9a=-18$ $a=2$ Thus, $f(x)=2(x-2)(x-4)^{2}$ $=(2x-4)(x^{2}-8x+16 )$ $=2x\left(x^{2}-8x+16\right)-4\left(x^{2}-8x+16\right)$ $=2x^{3}-16x^{2}+32x-4x^{2}+32x-64$ $f(x)=2x^{3}-20x^{2}+64x-64$
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