Divide using synthetic division $$ ( x^{5}+2x^{4}+x^{3}-36x ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&2&1&0&-36&0\\& & -3& 3& -12& 36& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{4}&\color{blue}{-12}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{5}+2x^{4}+x^{3}-36x }{ x+3 } = \color{blue}{x^{4}-x^{3}+4x^{2}-12x} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&2&1&0&-36&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&2&1&0&-36&0\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&2&1&0&-36&0\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ 2 }&1&0&-36&0\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{-1}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&2&1&0&-36&0\\& & -3& \color{blue}{3} & & & \\ \hline &1&\color{blue}{-1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 3 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-3&1&2&\color{orangered}{ 1 }&0&-36&0\\& & -3& \color{orangered}{3} & & & \\ \hline &1&-1&\color{orangered}{4}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&2&1&0&-36&0\\& & -3& 3& \color{blue}{-12} & & \\ \hline &1&-1&\color{blue}{4}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}-3&1&2&1&\color{orangered}{ 0 }&-36&0\\& & -3& 3& \color{orangered}{-12} & & \\ \hline &1&-1&4&\color{orangered}{-12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&2&1&0&-36&0\\& & -3& 3& -12& \color{blue}{36} & \\ \hline &1&-1&4&\color{blue}{-12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -36 } + \color{orangered}{ 36 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-3&1&2&1&0&\color{orangered}{ -36 }&0\\& & -3& 3& -12& \color{orangered}{36} & \\ \hline &1&-1&4&-12&\color{orangered}{0}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&2&1&0&-36&0\\& & -3& 3& -12& 36& \color{blue}{0} \\ \hline &1&-1&4&-12&\color{blue}{0}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-3&1&2&1&0&-36&\color{orangered}{ 0 }\\& & -3& 3& -12& 36& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{4}&\color{blue}{-12}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-x^{3}+4x^{2}-12x } $ with a remainder of $ \color{red}{ 0 } $.