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