## Intermediate Algebra (12th Edition)

$5p^{2}-11p+14$
Setup: (the variable is $p$ 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. $\begin{array}{rrr} {-1\ \ )} &{5}&{-6}&{3}&{14}&{ }\\ { } &{ }&{ }&{ }&{ } &{ }\\ \hline &{5 }&{ }&{ }&{ } &{ }\end{array}$ Fill the next entries, column by column: Middle row: k$\times$(previous bottom row entry) Bottom row: add the two entries above. Repeat. $\begin{array}{rrr} {-1\ \ )} &{5}&{-6}&{3}&{14}&{ }\\ { } &{ }&{-5}&{ 11}&{ -14} &{ }\\ \hline &{5 }&{ -11}&{14 }&{ 0} &{ }\end{array}$ Interpret result: Q(x) is the quotient, R(x) the remainder. $\displaystyle \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$ $Q(p)=5p^{2}-11p+14\quad R(p)=0$ $\displaystyle \frac{5p^{3}-6p^{2}+3p+14}{p+1}= 5p^{2}-11p+14$