Divide using synthetic division $$ ( x^{4}-3x^{3}-11x^{2}+3x+10 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&-3&-11&3&10\\& & 5& 10& -5& \color{black}{-10} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-3x^{3}-11x^{2}+3x+10 }{ x-5 } = \color{blue}{x^{3}+2x^{2}-x-2} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-3&-11&3&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&-3&-11&3&10\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-3&-11&3&10\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 5 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ -3 }&-11&3&10\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-3&-11&3&10\\& & 5& \color{blue}{10} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 10 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}5&1&-3&\color{orangered}{ -11 }&3&10\\& & 5& \color{orangered}{10} & & \\ \hline &1&2&\color{orangered}{-1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-3&-11&3&10\\& & 5& 10& \color{blue}{-5} & \\ \hline &1&2&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}5&1&-3&-11&\color{orangered}{ 3 }&10\\& & 5& 10& \color{orangered}{-5} & \\ \hline &1&2&-1&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-3&-11&3&10\\& & 5& 10& -5& \color{blue}{-10} \\ \hline &1&2&-1&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&1&-3&-11&3&\color{orangered}{ 10 }\\& & 5& 10& -5& \color{orangered}{-10} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}-x-2 } $ with a remainder of $ \color{red}{ 0 } $.