College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.5 - Polynomials - R.4 Assess Your Understanding - Page 49: 100

Answer

Quotient = $\quad x^{2}-\displaystyle \frac{2}{3}x-\frac{1}{9}$ Remainder = $\quad \displaystyle \frac{16}{9}x-\frac{17}{9}$

Work Step by Step

$\begin{array}{lllllll} & x^{2} & -\frac{2}{3}x & -\frac{1}{9} & & & \\ & -- & -- & -- & -- & & \\ 3x^{2}+x+1 \ \ ) & 3x^{4} & -x^{3} & +0 & +x & -2 & \\ & 3x^{4} & +x^{3} & +x^{2} & & & \\ & -- & -- & -- & -- & -- & \\ & & -2x^{3} & -x^{2} & +x & -2 & \\ & & -2x^{3} & -\frac{2}{3}x^{2} & -\frac{2}{3}x & & \\ & & -- & -- & -- & -- & \\ & & & -\frac{1}{3}x^{2} & +\frac{5}{3}x & -2 & \\ & & & -\frac{1}{3}x^{2} & -\frac{1}{9}x & -\frac{1}{9} & \\ & & & -- & -- & -- & \\ & & & & \frac{16}{9}x & -\frac{17}{9} & \\ & & & & & & \end{array}$ Check: $(3x^{2}+x+1)(x^{2}-\displaystyle \frac{2}{3}x-\frac{1}{9})+\frac{16}{9}x-\frac{17}{9}$ $=3x^{4}-2x^{3}-\displaystyle \frac{1}{3}x^{2} $ $\qquad \ \ \ \ +x^{3}-\frac{2}{3}x^{2}- \frac{1}{9}x\ $ $\ \ \qquad\qquad++x^{2}-\frac{2}{3}x-\frac{1}{9}+\frac{16}{9}x-\frac{17}{9}$ $=3x^{4}-x^{3}+x-2$
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