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 - Section 3.2 The Real Zeros of a Polynomial Function - 3.2 Assess Your Understanding - Page 226: 121

Answer

$\dfrac{1}{3}$ is not a zero of the function by the factor theorem.

Work Step by Step

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 are given the function $f(x)=4x^3-5x^2-3x+1$ We simplify the given equation as follows: $f(\dfrac{1}{3})=4(\dfrac{1}{3})^3-5(\dfrac{1}{3})^2-3(\dfrac{1}{3})+1 \\=\dfrac{-11}{27}$ This implies that $f(\dfrac{1}{3}) \ne 0$ So, $\dfrac{1}{3}$ is not a zero of the function by the factor theorem.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.