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 363: 23

Answer

quotient $6x^4+12x^3+22x^2+48x+93$ and remainder $r(x)=\frac{187}{x-2}$
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Work Step by Step

Step 1. The coefficients of the dividend can be identified as $\{6,0,-2,4 -3,1\}$ and the divisor as $x-2$; use synthetic division as shown in the figure to get the quotient and the remainder. Step 2. We can identify the result as $\frac{6x^5-2x^3+4x^2-3x+1}{x-2}=6x^4+12x^3+22x^2+48x+93+\frac{187}{x-2}$ with the quotient as $6x^4+12x^3+22x^2+48x+93$ and the remainder as $r(x)=\frac{187}{x-2}$
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