Divide using synthetic division $$ ( -4x^{3}+7x^{2}+12x+9 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-4&7&12&9\\& & -12& -15& \color{black}{-9} \\ \hline &\color{blue}{-4}&\color{blue}{-5}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -4x^{3}+7x^{2}+12x+9 }{ x-3 } = \color{blue}{-4x^{2}-5x-3} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&7&12&9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -4 }&7&12&9\\& & & & \\ \hline &\color{orangered}{-4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&7&12&9\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{-4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}3&-4&\color{orangered}{ 7 }&12&9\\& & \color{orangered}{-12} & & \\ \hline &-4&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&7&12&9\\& & -12& \color{blue}{-15} & \\ \hline &-4&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}3&-4&7&\color{orangered}{ 12 }&9\\& & -12& \color{orangered}{-15} & \\ \hline &-4&-5&\color{orangered}{-3}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&7&12&9\\& & -12& -15& \color{blue}{-9} \\ \hline &-4&-5&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&-4&7&12&\color{orangered}{ 9 }\\& & -12& -15& \color{orangered}{-9} \\ \hline &\color{blue}{-4}&\color{blue}{-5}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -4x^{2}-5x-3 } $ with a remainder of $ \color{red}{ 0 } $.