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: 61

Answer

$f(x)=x^{3}-3x^{2}+x+1$ (sample answer with $a=1$)

Work Step by Step

If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem) So, $(x-1)$ is a factor of f, $(x-(1+\sqrt{2}))=[(x-1)-\sqrt{2}]$ is a factor of f, $(x-(1-\sqrt{2}))=[(x-1)+\sqrt{2}]$ is a factor of f. We assume multiplicity 1, so we have $f(x)=a(x-1)[(x-1)-\sqrt{2}] [(x-1)+\sqrt{2}]$ We are asked for $a$ polynomial, so we choose the one for which $a=1.$ $f(x)=(x-1)[(x-1)-\sqrt{2}] [(x-1)+\sqrt{2}]\quad$... Recognize a difference of squares: $f(x)=(x-1)[(x-1)^2-(\sqrt{2})^{2}]$ $f(x)=(x-1)[x^{2}-2x+1-2]$ $ f(x)=(x-1)(x^{2}-2x-1)\quad$... distribute, $f(x)=x(x^{2}-2x-1)-(x^{2}-2x-1)$ $f(x)=x^{3}-2x^{2}-x-x^{2}+2x+1$ $f(x)=x^{3}-3x^{2}+x+1$
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.