Divide using synthetic division $$ ( 3x^{3}-16x^{2}-72 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}6&3&-16&0&-72\\& & 18& 12& \color{black}{72} \\ \hline &\color{blue}{3}&\color{blue}{2}&\color{blue}{12}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-16x^{2}-72 }{ x-6 } = \color{blue}{3x^{2}+2x+12} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{6}&3&-16&0&-72\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}6&\color{orangered}{ 3 }&-16&0&-72\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 3 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&3&-16&0&-72\\& & \color{blue}{18} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 18 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}6&3&\color{orangered}{ -16 }&0&-72\\& & \color{orangered}{18} & & \\ \hline &3&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 2 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&3&-16&0&-72\\& & 18& \color{blue}{12} & \\ \hline &3&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}6&3&-16&\color{orangered}{ 0 }&-72\\& & 18& \color{orangered}{12} & \\ \hline &3&2&\color{orangered}{12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 12 } = \color{blue}{ 72 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&3&-16&0&-72\\& & 18& 12& \color{blue}{72} \\ \hline &3&2&\color{blue}{12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -72 } + \color{orangered}{ 72 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}6&3&-16&0&\color{orangered}{ -72 }\\& & 18& 12& \color{orangered}{72} \\ \hline &\color{blue}{3}&\color{blue}{2}&\color{blue}{12}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+2x+12 } $ with a remainder of $ \color{red}{ 0 } $.