Precalculus (6th Edition) Blitzer

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

Chapter 2 - Review Exercises - Page 431: 62


$f(x)=2x^4+12x^3+20x^2+12x+18$ See graph.

Work Step by Step

Step 1. Given a zero $x=i$, we can identify another zero $x=-i$ Step 2. With a zero $x=-3$ of multiplicity of 2, we have $(x+3)^2$ as a factor. Step 3. For $n=4$, we can write the polynomial as $f(x)=(x-i)(x+i)(x+3)^2=(x^2+1)(x^2+6x+9)=x^4+6x^3+10x^2+6x+9$ Step 4. Let $x=-1$; we have $f(-1)=(-1)^4+6(-1)^3+10(-1)^2+6(-1)+9=1-6+10-6+9=8$ Thus, we are missing a factor of $2$ for the polynomial. Step 5. We conclude the polynomial should be $f(x)=2x^4+12x^3+20x^2+12x+18$ Step 6. See graph for the zeros and $f(-1)$
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