Divide using synthetic division $$ ( 3x^{3}-11x^{2}-12 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&3&-11&0&-12\\& & 12& 4& \color{black}{16} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{4}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-11x^{2}-12 }{ x-4 } = \color{blue}{3x^{2}+x+4} ~+~ \frac{ \color{red}{ 4 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-11&0&-12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 3 }&-11&0&-12\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-11&0&-12\\& & \color{blue}{12} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 12 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}4&3&\color{orangered}{ -11 }&0&-12\\& & \color{orangered}{12} & & \\ \hline &3&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-11&0&-12\\& & 12& \color{blue}{4} & \\ \hline &3&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}4&3&-11&\color{orangered}{ 0 }&-12\\& & 12& \color{orangered}{4} & \\ \hline &3&1&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 4 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-11&0&-12\\& & 12& 4& \color{blue}{16} \\ \hline &3&1&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 16 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}4&3&-11&0&\color{orangered}{ -12 }\\& & 12& 4& \color{orangered}{16} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{4}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+x+4 } $ with a remainder of $ \color{red}{ 4 } $.