Divide using synthetic division $$ ( x^{4}+2x^{3}+x^{2}-10x ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&2&1&-10&0\\& & 3& 15& 48& \color{black}{114} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{16}&\color{blue}{38}&\color{orangered}{114} \end{array} $$The solution is:
$$ \frac{ x^{4}+2x^{3}+x^{2}-10x }{ x-3 } = \color{blue}{x^{3}+5x^{2}+16x+38} ~+~ \frac{ \color{red}{ 114 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&2&1&-10&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&2&1&-10&0\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&2&1&-10&0\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 3 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ 2 }&1&-10&0\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&2&1&-10&0\\& & 3& \color{blue}{15} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 15 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrr}3&1&2&\color{orangered}{ 1 }&-10&0\\& & 3& \color{orangered}{15} & & \\ \hline &1&5&\color{orangered}{16}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&2&1&-10&0\\& & 3& 15& \color{blue}{48} & \\ \hline &1&5&\color{blue}{16}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 48 } = \color{orangered}{ 38 } $
$$ \begin{array}{c|rrrrr}3&1&2&1&\color{orangered}{ -10 }&0\\& & 3& 15& \color{orangered}{48} & \\ \hline &1&5&16&\color{orangered}{38}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 38 } = \color{blue}{ 114 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&2&1&-10&0\\& & 3& 15& 48& \color{blue}{114} \\ \hline &1&5&16&\color{blue}{38}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 114 } = \color{orangered}{ 114 } $
$$ \begin{array}{c|rrrrr}3&1&2&1&-10&\color{orangered}{ 0 }\\& & 3& 15& 48& \color{orangered}{114} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{16}&\color{blue}{38}&\color{orangered}{114} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}+16x+38 } $ with a remainder of $ \color{red}{ 114 } $.