Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.4 - Dividing Polynomials; Remainder and Factor Theorems - Exercise Set - Page 364: 51

Answer

a. See figure and explanations. b. $3\ mm$
1581975883

Work Step by Step

a. The coefficients of the dividend $f(x)=14x^3-17x^2-16x-177$ can be identified in order as $\{14,-17,-16,-177\}$ and the divisor is $x-3$; using synthetic division as shown in the figure, we get the remainder as zero, indicating that $x=3$ is a solution of the equation. b. For this function, let $f(x)=211$, we have $14x^3-17x^2-16x+34=211$ and $14x^3-17x^2-16x-177=0$. Based on the above result, we have $(14x^2+25x+59)(x-3)=0$. Since the only real solution to the equation is $x=3$, we have the abdominal width as $3\ mm$.
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.