Divide using synthetic division $$ ( x^{4}-6x^{3}-33x^{2}+46x ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&-6&-33&46&0\\& & 3& -9& -126& \color{black}{-240} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-42}&\color{blue}{-80}&\color{orangered}{-240} \end{array} $$The solution is:
$$ \frac{ x^{4}-6x^{3}-33x^{2}+46x }{ x-3 } = \color{blue}{x^{3}-3x^{2}-42x-80} \color{red}{~-~} \frac{ \color{red}{ 240 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-6&-33&46&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&-6&-33&46&0\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-6&-33&46&0\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 3 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ -6 }&-33&46&0\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-6&-33&46&0\\& & 3& \color{blue}{-9} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -33 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -42 } $
$$ \begin{array}{c|rrrrr}3&1&-6&\color{orangered}{ -33 }&46&0\\& & 3& \color{orangered}{-9} & & \\ \hline &1&-3&\color{orangered}{-42}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -42 \right) } = \color{blue}{ -126 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-6&-33&46&0\\& & 3& -9& \color{blue}{-126} & \\ \hline &1&-3&\color{blue}{-42}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 46 } + \color{orangered}{ \left( -126 \right) } = \color{orangered}{ -80 } $
$$ \begin{array}{c|rrrrr}3&1&-6&-33&\color{orangered}{ 46 }&0\\& & 3& -9& \color{orangered}{-126} & \\ \hline &1&-3&-42&\color{orangered}{-80}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -80 \right) } = \color{blue}{ -240 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-6&-33&46&0\\& & 3& -9& -126& \color{blue}{-240} \\ \hline &1&-3&-42&\color{blue}{-80}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -240 \right) } = \color{orangered}{ -240 } $
$$ \begin{array}{c|rrrrr}3&1&-6&-33&46&\color{orangered}{ 0 }\\& & 3& -9& -126& \color{orangered}{-240} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-42}&\color{blue}{-80}&\color{orangered}{-240} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3x^{2}-42x-80 } $ with a remainder of $ \color{red}{ -240 } $.