Divide using synthetic division $$ ( -6x^{3}-x ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&-6&0&-1&0\\& & 12& -24& \color{black}{50} \\ \hline &\color{blue}{-6}&\color{blue}{12}&\color{blue}{-25}&\color{orangered}{50} \end{array} $$The solution is:
$$ \frac{ -6x^{3}-x }{ x+2 } = \color{blue}{-6x^{2}+12x-25} ~+~ \frac{ \color{red}{ 50 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-6&0&-1&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ -6 }&0&-1&0\\& & & & \\ \hline &\color{orangered}{-6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-6&0&-1&0\\& & \color{blue}{12} & & \\ \hline &\color{blue}{-6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}-2&-6&\color{orangered}{ 0 }&-1&0\\& & \color{orangered}{12} & & \\ \hline &-6&\color{orangered}{12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 12 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-6&0&-1&0\\& & 12& \color{blue}{-24} & \\ \hline &-6&\color{blue}{12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -25 } $
$$ \begin{array}{c|rrrr}-2&-6&0&\color{orangered}{ -1 }&0\\& & 12& \color{orangered}{-24} & \\ \hline &-6&12&\color{orangered}{-25}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -25 \right) } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-6&0&-1&0\\& & 12& -24& \color{blue}{50} \\ \hline &-6&12&\color{blue}{-25}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 50 } = \color{orangered}{ 50 } $
$$ \begin{array}{c|rrrr}-2&-6&0&-1&\color{orangered}{ 0 }\\& & 12& -24& \color{orangered}{50} \\ \hline &\color{blue}{-6}&\color{blue}{12}&\color{blue}{-25}&\color{orangered}{50} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -6x^{2}+12x-25 } $ with a remainder of $ \color{red}{ 50 } $.