Divide using synthetic division $$ ( 12x^{3}+10x^{2}+5x+1 ) \div ( 3x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&12&10&5&1\\& & -12& 2& \color{black}{-7} \\ \hline &\color{blue}{12}&\color{blue}{-2}&\color{blue}{7}&\color{orangered}{-6} \end{array} $$The solution is:
$$ \frac{ 12x^{3}+10x^{2}+5x+1 }{ x+1 } = \color{blue}{12x^{2}-2x+7} \color{red}{~-~} \frac{ \color{red}{ 6 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&12&10&5&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 12 }&10&5&1\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 12 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&12&10&5&1\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-1&12&\color{orangered}{ 10 }&5&1\\& & \color{orangered}{-12} & & \\ \hline &12&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&12&10&5&1\\& & -12& \color{blue}{2} & \\ \hline &12&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 2 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-1&12&10&\color{orangered}{ 5 }&1\\& & -12& \color{orangered}{2} & \\ \hline &12&-2&\color{orangered}{7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 7 } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&12&10&5&1\\& & -12& 2& \color{blue}{-7} \\ \hline &12&-2&\color{blue}{7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-1&12&10&5&\color{orangered}{ 1 }\\& & -12& 2& \color{orangered}{-7} \\ \hline &\color{blue}{12}&\color{blue}{-2}&\color{blue}{7}&\color{orangered}{-6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{2}-2x+7 } $ with a remainder of $ \color{red}{ -6 } $.