Divide using synthetic division $$ ( 5x^{3}-12x^{2}+11x+12 ) \div ( 5x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&5&-12&11&12\\& & -15& 81& \color{black}{-276} \\ \hline &\color{blue}{5}&\color{blue}{-27}&\color{blue}{92}&\color{orangered}{-264} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-12x^{2}+11x+12 }{ x+3 } = \color{blue}{5x^{2}-27x+92} \color{red}{~-~} \frac{ \color{red}{ 264 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-12&11&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 5 }&-12&11&12\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-12&11&12\\& & \color{blue}{-15} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -27 } $
$$ \begin{array}{c|rrrr}-3&5&\color{orangered}{ -12 }&11&12\\& & \color{orangered}{-15} & & \\ \hline &5&\color{orangered}{-27}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -27 \right) } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-12&11&12\\& & -15& \color{blue}{81} & \\ \hline &5&\color{blue}{-27}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 81 } = \color{orangered}{ 92 } $
$$ \begin{array}{c|rrrr}-3&5&-12&\color{orangered}{ 11 }&12\\& & -15& \color{orangered}{81} & \\ \hline &5&-27&\color{orangered}{92}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 92 } = \color{blue}{ -276 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-12&11&12\\& & -15& 81& \color{blue}{-276} \\ \hline &5&-27&\color{blue}{92}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -276 \right) } = \color{orangered}{ -264 } $
$$ \begin{array}{c|rrrr}-3&5&-12&11&\color{orangered}{ 12 }\\& & -15& 81& \color{orangered}{-276} \\ \hline &\color{blue}{5}&\color{blue}{-27}&\color{blue}{92}&\color{orangered}{-264} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-27x+92 } $ with a remainder of $ \color{red}{ -264 } $.