Divide using synthetic division $$ ( x^{3}-12x^{2}+33x ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&-12&33&0\\& & 5& -35& \color{black}{-10} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{-2}&\color{orangered}{-10} \end{array} $$The solution is:
$$ \frac{ x^{3}-12x^{2}+33x }{ x-5 } = \color{blue}{x^{2}-7x-2} \color{red}{~-~} \frac{ \color{red}{ 10 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-12&33&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 1 }&-12&33&0\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-12&33&0\\& & \color{blue}{5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 5 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ -12 }&33&0\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -35 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-12&33&0\\& & 5& \color{blue}{-35} & \\ \hline &1&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 33 } + \color{orangered}{ \left( -35 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}5&1&-12&\color{orangered}{ 33 }&0\\& & 5& \color{orangered}{-35} & \\ \hline &1&-7&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-12&33&0\\& & 5& -35& \color{blue}{-10} \\ \hline &1&-7&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}5&1&-12&33&\color{orangered}{ 0 }\\& & 5& -35& \color{orangered}{-10} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{-2}&\color{orangered}{-10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-7x-2 } $ with a remainder of $ \color{red}{ -10 } $.