Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 3 - Polynomial and Rational Functions - Chapter Review - Review Exercises - Page 268: 30

Answer

The complex zeros are: $\dfrac{1}{2}$ with multiplicity $2$, $-2$ with multiplicity $1$. The function is: $f(x)=4(x-\dfrac{1}{2})^2 (x+2)$

Work Step by Step

We are given the function as: $f(x)=4x^3+4x^2-7x+2$ The factor theorem states that when $f(a)=0$, then we have $(x-a)$ as a factor of $f(x)$ and when $(x-a)$ is a factor of $f(x)$, then $f(a)=0$. We can see that $f(-2)=0$. This means that $x+2$ is a factor of $f(x)=4x^3+4x^2-7x+2$ . Simplify the given function as follows: $4x^3+4x^2-7x+2 \\ (x+2)(4x^2-4x+1)=0 \\ (x+2)(2x-1)(2x-1)=0$ By the zero product property, we have: $x=\dfrac{1}{2}, \dfrac{1}{2}$ Thus, we have: The complex zeros are: $\dfrac{1}{2}$ with multiplicity $2$, $-2$ with multiplicity $1$. The function is: $f(x)=4(x-\dfrac{1}{2})^2 (x+2)$
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