Divide using synthetic division $$ ( x^{3}+3x^{2}-36x+32 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&3&-36&32\\& & 4& 28& \color{black}{-32} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}+3x^{2}-36x+32 }{ x-4 } = \color{blue}{x^{2}+7x-8} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&3&-36&32\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&3&-36&32\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&3&-36&32\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 4 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ 3 }&-36&32\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 7 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&3&-36&32\\& & 4& \color{blue}{28} & \\ \hline &1&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -36 } + \color{orangered}{ 28 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}4&1&3&\color{orangered}{ -36 }&32\\& & 4& \color{orangered}{28} & \\ \hline &1&7&\color{orangered}{-8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&3&-36&32\\& & 4& 28& \color{blue}{-32} \\ \hline &1&7&\color{blue}{-8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 32 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}4&1&3&-36&\color{orangered}{ 32 }\\& & 4& 28& \color{orangered}{-32} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+7x-8 } $ with a remainder of $ \color{red}{ 0 } $.