Divide using synthetic division $$ ( x^{4}-2x^{3}+x^{2}+12x-6 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&-2&1&12&-6\\& & -4& 24& -100& \color{black}{352} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{25}&\color{blue}{-88}&\color{orangered}{346} \end{array} $$The solution is:
$$ \frac{ x^{4}-2x^{3}+x^{2}+12x-6 }{ x+4 } = \color{blue}{x^{3}-6x^{2}+25x-88} ~+~ \frac{ \color{red}{ 346 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-2&1&12&-6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&-2&1&12&-6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-2&1&12&-6\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ -2 }&1&12&-6\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-2&1&12&-6\\& & -4& \color{blue}{24} & & \\ \hline &1&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 24 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrrr}-4&1&-2&\color{orangered}{ 1 }&12&-6\\& & -4& \color{orangered}{24} & & \\ \hline &1&-6&\color{orangered}{25}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 25 } = \color{blue}{ -100 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-2&1&12&-6\\& & -4& 24& \color{blue}{-100} & \\ \hline &1&-6&\color{blue}{25}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -100 \right) } = \color{orangered}{ -88 } $
$$ \begin{array}{c|rrrrr}-4&1&-2&1&\color{orangered}{ 12 }&-6\\& & -4& 24& \color{orangered}{-100} & \\ \hline &1&-6&25&\color{orangered}{-88}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -88 \right) } = \color{blue}{ 352 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-2&1&12&-6\\& & -4& 24& -100& \color{blue}{352} \\ \hline &1&-6&25&\color{blue}{-88}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 352 } = \color{orangered}{ 346 } $
$$ \begin{array}{c|rrrrr}-4&1&-2&1&12&\color{orangered}{ -6 }\\& & -4& 24& -100& \color{orangered}{352} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{25}&\color{blue}{-88}&\color{orangered}{346} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-6x^{2}+25x-88 } $ with a remainder of $ \color{red}{ 346 } $.