Divide using synthetic division $$ ( 3x^{4}-6x^{3}-39x^{2}-30x ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&3&-6&-39&-30&0\\& & 15& 45& 30& \color{black}{0} \\ \hline &\color{blue}{3}&\color{blue}{9}&\color{blue}{6}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-6x^{3}-39x^{2}-30x }{ x-5 } = \color{blue}{3x^{3}+9x^{2}+6x} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-6&-39&-30&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 3 }&-6&-39&-30&0\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-6&-39&-30&0\\& & \color{blue}{15} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 15 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}5&3&\color{orangered}{ -6 }&-39&-30&0\\& & \color{orangered}{15} & & & \\ \hline &3&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 9 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-6&-39&-30&0\\& & 15& \color{blue}{45} & & \\ \hline &3&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -39 } + \color{orangered}{ 45 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}5&3&-6&\color{orangered}{ -39 }&-30&0\\& & 15& \color{orangered}{45} & & \\ \hline &3&9&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 6 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-6&-39&-30&0\\& & 15& 45& \color{blue}{30} & \\ \hline &3&9&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 30 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&3&-6&-39&\color{orangered}{ -30 }&0\\& & 15& 45& \color{orangered}{30} & \\ \hline &3&9&6&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-6&-39&-30&0\\& & 15& 45& 30& \color{blue}{0} \\ \hline &3&9&6&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&3&-6&-39&-30&\color{orangered}{ 0 }\\& & 15& 45& 30& \color{orangered}{0} \\ \hline &\color{blue}{3}&\color{blue}{9}&\color{blue}{6}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+9x^{2}+6x } $ with a remainder of $ \color{red}{ 0 } $.