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