Divide using synthetic division $$ ( 4x^{3}+4x^{2}-9x-9 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&4&-9&-9\\& & 12& 48& \color{black}{117} \\ \hline &\color{blue}{4}&\color{blue}{16}&\color{blue}{39}&\color{orangered}{108} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+4x^{2}-9x-9 }{ x-3 } = \color{blue}{4x^{2}+16x+39} ~+~ \frac{ \color{red}{ 108 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&4&-9&-9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&4&-9&-9\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&4&-9&-9\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 12 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ 4 }&-9&-9\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&4&-9&-9\\& & 12& \color{blue}{48} & \\ \hline &4&\color{blue}{16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 48 } = \color{orangered}{ 39 } $
$$ \begin{array}{c|rrrr}3&4&4&\color{orangered}{ -9 }&-9\\& & 12& \color{orangered}{48} & \\ \hline &4&16&\color{orangered}{39}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 39 } = \color{blue}{ 117 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&4&-9&-9\\& & 12& 48& \color{blue}{117} \\ \hline &4&16&\color{blue}{39}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 117 } = \color{orangered}{ 108 } $
$$ \begin{array}{c|rrrr}3&4&4&-9&\color{orangered}{ -9 }\\& & 12& 48& \color{orangered}{117} \\ \hline &\color{blue}{4}&\color{blue}{16}&\color{blue}{39}&\color{orangered}{108} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+16x+39 } $ with a remainder of $ \color{red}{ 108 } $.