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