Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Appendix - Synthetic Division - Exercises - Page 646: 10

Answer

$z+9+\dfrac{39}{z-5}$

Work Step by Step

Setup: (the variable is $z$ instead of $x$) Dividing with $x-k$, place k as the top left entry. List coefficients of the numerator (0 for missing powers) in the first row. Copy the leading coefficient to the bottom row, same column. $\begin{array}{rrr} {5\ \ )} &{1}&{4}&{-6}\\ { } &{ }&{ } &{ }\\ \hline &{1 }&{ } &{ }\end{array}$ Fill the next entries, column by column: Middle row: k$\times$(previous bottom row entry) Bottom row: add the two entries above. Repeat. $\begin{array}{rrr} {5\ \ )} &{1}&{4}&{-6}\\ { } &{ }&{5 } &{45 }\\ \hline &{1 }&{ 9} &{ 39}\end{array}$ Interpret result: Q(x) is the quotient, R(x) the remainder. $\displaystyle \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$ $Q(z)=z+9\quad R(p)=39$ $\displaystyle \frac{z^{2}+4z-6}{z-5}=z+9+\frac{39}{z-5}$
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