Divide using synthetic division $$ ( -6x^{4}-5x^{2}+8 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&-6&0&-5&0&8\\& & 12& -24& 58& \color{black}{-116} \\ \hline &\color{blue}{-6}&\color{blue}{12}&\color{blue}{-29}&\color{blue}{58}&\color{orangered}{-108} \end{array} $$The solution is:
$$ \frac{ -6x^{4}-5x^{2}+8 }{ x+2 } = \color{blue}{-6x^{3}+12x^{2}-29x+58} \color{red}{~-~} \frac{ \color{red}{ 108 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-6&0&-5&0&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ -6 }&0&-5&0&8\\& & & & & \\ \hline &\color{orangered}{-6}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-6&0&-5&0&8\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{-6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-2&-6&\color{orangered}{ 0 }&-5&0&8\\& & \color{orangered}{12} & & & \\ \hline &-6&\color{orangered}{12}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-6&0&-5&0&8\\& & 12& \color{blue}{-24} & & \\ \hline &-6&\color{blue}{12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -29 } $
$$ \begin{array}{c|rrrrr}-2&-6&0&\color{orangered}{ -5 }&0&8\\& & 12& \color{orangered}{-24} & & \\ \hline &-6&12&\color{orangered}{-29}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -29 \right) } = \color{blue}{ 58 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-6&0&-5&0&8\\& & 12& -24& \color{blue}{58} & \\ \hline &-6&12&\color{blue}{-29}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 58 } = \color{orangered}{ 58 } $
$$ \begin{array}{c|rrrrr}-2&-6&0&-5&\color{orangered}{ 0 }&8\\& & 12& -24& \color{orangered}{58} & \\ \hline &-6&12&-29&\color{orangered}{58}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 58 } = \color{blue}{ -116 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-6&0&-5&0&8\\& & 12& -24& 58& \color{blue}{-116} \\ \hline &-6&12&-29&\color{blue}{58}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -116 \right) } = \color{orangered}{ -108 } $
$$ \begin{array}{c|rrrrr}-2&-6&0&-5&0&\color{orangered}{ 8 }\\& & 12& -24& 58& \color{orangered}{-116} \\ \hline &\color{blue}{-6}&\color{blue}{12}&\color{blue}{-29}&\color{blue}{58}&\color{orangered}{-108} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -6x^{3}+12x^{2}-29x+58 } $ with a remainder of $ \color{red}{ -108 } $.