Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.5 - Zeros of Polynomial Functions - Exercise Set - Page 378: 60

Answer

a. $x=-\frac{1}{5},1,\pm 2i$ b. See graph and explanations.
1582126402

Work Step by Step

a. Based on the graph, we can identify a zero as $x=1$. However, as the function is of $4th$ degree, we need to find another zero to reduce it to a quadratic form. Using synthetic division with possible zeros of $\pm1,\pm2,\pm4,\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5}$, we can find an additional zero at $x=-\frac{1}{5}$ and the quotient as $-5x^2-20=-5(x^2+4)$ (shown in the figure). Thus the zeros are $x=-\frac{1}{5},1,\pm 2i$ b. The end behaviors of the function can be found as $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$ . The maximum number of of turning points is $4-1=3$ (reduced to 1 due to imaginal zeros), and the y-intercept can be found as $y=f(0)=4$. With the above information, we can finish the graph as shown in the figure, where the locations and values of the extrema are not a real concern at this point.
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.