Divide using synthetic division $$ ( x^{4}+7x^{3}-44x+18 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&7&0&-44&18\\& & -4& -12& 48& \color{black}{-16} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-12}&\color{blue}{4}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ x^{4}+7x^{3}-44x+18 }{ x+4 } = \color{blue}{x^{3}+3x^{2}-12x+4} ~+~ \frac{ \color{red}{ 2 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&7&0&-44&18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&7&0&-44&18\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&7&0&-44&18\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ 7 }&0&-44&18\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 3 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&7&0&-44&18\\& & -4& \color{blue}{-12} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}-4&1&7&\color{orangered}{ 0 }&-44&18\\& & -4& \color{orangered}{-12} & & \\ \hline &1&3&\color{orangered}{-12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&7&0&-44&18\\& & -4& -12& \color{blue}{48} & \\ \hline &1&3&\color{blue}{-12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -44 } + \color{orangered}{ 48 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-4&1&7&0&\color{orangered}{ -44 }&18\\& & -4& -12& \color{orangered}{48} & \\ \hline &1&3&-12&\color{orangered}{4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&7&0&-44&18\\& & -4& -12& 48& \color{blue}{-16} \\ \hline &1&3&-12&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-4&1&7&0&-44&\color{orangered}{ 18 }\\& & -4& -12& 48& \color{orangered}{-16} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-12}&\color{blue}{4}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+3x^{2}-12x+4 } $ with a remainder of $ \color{red}{ 2 } $.