Divide using synthetic division $$ ( x^{3}-7x^{2}+33 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-7&0&33\\& & 3& -12& \color{black}{-36} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-12}&\color{orangered}{-3} \end{array} $$The solution is:
$$ \frac{ x^{3}-7x^{2}+33 }{ x-3 } = \color{blue}{x^{2}-4x-12} \color{red}{~-~} \frac{ \color{red}{ 3 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-7&0&33\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-7&0&33\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-7&0&33\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 3 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -7 }&0&33\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-7&0&33\\& & 3& \color{blue}{-12} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}3&1&-7&\color{orangered}{ 0 }&33\\& & 3& \color{orangered}{-12} & \\ \hline &1&-4&\color{orangered}{-12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-7&0&33\\& & 3& -12& \color{blue}{-36} \\ \hline &1&-4&\color{blue}{-12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 33 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}3&1&-7&0&\color{orangered}{ 33 }\\& & 3& -12& \color{orangered}{-36} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-12}&\color{orangered}{-3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-4x-12 } $ with a remainder of $ \color{red}{ -3 } $.