Divide using synthetic division $$ ( 2x^{3}-8x^{2}-19x-30 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}6&2&-8&-19&-30\\& & 12& 24& \color{black}{30} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-8x^{2}-19x-30 }{ x-6 } = \color{blue}{2x^{2}+4x+5} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&-8&-19&-30\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}6&\color{orangered}{ 2 }&-8&-19&-30\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 2 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&-8&-19&-30\\& & \color{blue}{12} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 12 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}6&2&\color{orangered}{ -8 }&-19&-30\\& & \color{orangered}{12} & & \\ \hline &2&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 4 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&-8&-19&-30\\& & 12& \color{blue}{24} & \\ \hline &2&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 24 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}6&2&-8&\color{orangered}{ -19 }&-30\\& & 12& \color{orangered}{24} & \\ \hline &2&4&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 5 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&-8&-19&-30\\& & 12& 24& \color{blue}{30} \\ \hline &2&4&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 30 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}6&2&-8&-19&\color{orangered}{ -30 }\\& & 12& 24& \color{orangered}{30} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+4x+5 } $ with a remainder of $ \color{red}{ 0 } $.