Divide using synthetic division $$ ( x^{4}+x^{3}+47x-2 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&1&0&47&-2\\& & -4& 12& -48& \color{black}{4} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{12}&\color{blue}{-1}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ x^{4}+x^{3}+47x-2 }{ x+4 } = \color{blue}{x^{3}-3x^{2}+12x-1} ~+~ \frac{ \color{red}{ 2 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&0&47&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&1&0&47&-2\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&0&47&-2\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ 1 }&0&47&-2\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&0&47&-2\\& & -4& \color{blue}{12} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-4&1&1&\color{orangered}{ 0 }&47&-2\\& & -4& \color{orangered}{12} & & \\ \hline &1&-3&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 12 } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&0&47&-2\\& & -4& 12& \color{blue}{-48} & \\ \hline &1&-3&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 47 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-4&1&1&0&\color{orangered}{ 47 }&-2\\& & -4& 12& \color{orangered}{-48} & \\ \hline &1&-3&12&\color{orangered}{-1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&0&47&-2\\& & -4& 12& -48& \color{blue}{4} \\ \hline &1&-3&12&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 4 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-4&1&1&0&47&\color{orangered}{ -2 }\\& & -4& 12& -48& \color{orangered}{4} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{12}&\color{blue}{-1}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3x^{2}+12x-1 } $ with a remainder of $ \color{red}{ 2 } $.