Divide using synthetic division $$ ( x^{3}+9x^{2}+x-10 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&9&1&-10\\& & 4& 52& \color{black}{212} \\ \hline &\color{blue}{1}&\color{blue}{13}&\color{blue}{53}&\color{orangered}{202} \end{array} $$The solution is:
$$ \frac{ x^{3}+9x^{2}+x-10 }{ x-4 } = \color{blue}{x^{2}+13x+53} ~+~ \frac{ \color{red}{ 202 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&9&1&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&9&1&-10\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&9&1&-10\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 4 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ 9 }&1&-10\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 13 } = \color{blue}{ 52 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&9&1&-10\\& & 4& \color{blue}{52} & \\ \hline &1&\color{blue}{13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 52 } = \color{orangered}{ 53 } $
$$ \begin{array}{c|rrrr}4&1&9&\color{orangered}{ 1 }&-10\\& & 4& \color{orangered}{52} & \\ \hline &1&13&\color{orangered}{53}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 53 } = \color{blue}{ 212 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&9&1&-10\\& & 4& 52& \color{blue}{212} \\ \hline &1&13&\color{blue}{53}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 212 } = \color{orangered}{ 202 } $
$$ \begin{array}{c|rrrr}4&1&9&1&\color{orangered}{ -10 }\\& & 4& 52& \color{orangered}{212} \\ \hline &\color{blue}{1}&\color{blue}{13}&\color{blue}{53}&\color{orangered}{202} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+13x+53 } $ with a remainder of $ \color{red}{ 202 } $.