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 430: 43

Answer

quotient $2x^2+3x-1$, remainder $0$
1582496056

Work Step by Step

Step 1. Using long division as shown in the figure, we have $\frac{4x^4+6x^3+3x-1}{2x^2+1}=2x^2+3x-1+\frac{0}{5x-3}$ Step 2. We can identify the quotient as $2x^2+3x-1$ and the remainder as $0$
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