Divide using synthetic division $$ ( x^{4}+x^{3}-4x^{2} ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&1&-4&0&0\\& & -4& 12& -32& \color{black}{128} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{8}&\color{blue}{-32}&\color{orangered}{128} \end{array} $$The solution is:
$$ \frac{ x^{4}+x^{3}-4x^{2} }{ x+4 } = \color{blue}{x^{3}-3x^{2}+8x-32} ~+~ \frac{ \color{red}{ 128 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&-4&0&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&1&-4&0&0\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&-4&0&0\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ 1 }&-4&0&0\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&-4&0&0\\& & -4& \color{blue}{12} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 12 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-4&1&1&\color{orangered}{ -4 }&0&0\\& & -4& \color{orangered}{12} & & \\ \hline &1&-3&\color{orangered}{8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 8 } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&-4&0&0\\& & -4& 12& \color{blue}{-32} & \\ \hline &1&-3&\color{blue}{8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -32 } $
$$ \begin{array}{c|rrrrr}-4&1&1&-4&\color{orangered}{ 0 }&0\\& & -4& 12& \color{orangered}{-32} & \\ \hline &1&-3&8&\color{orangered}{-32}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -32 \right) } = \color{blue}{ 128 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&-4&0&0\\& & -4& 12& -32& \color{blue}{128} \\ \hline &1&-3&8&\color{blue}{-32}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 128 } = \color{orangered}{ 128 } $
$$ \begin{array}{c|rrrrr}-4&1&1&-4&0&\color{orangered}{ 0 }\\& & -4& 12& -32& \color{orangered}{128} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{8}&\color{blue}{-32}&\color{orangered}{128} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3x^{2}+8x-32 } $ with a remainder of $ \color{red}{ 128 } $.