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