Divide using synthetic division $$ ( -2x^{3}-8x^{2}+12x+10 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&-2&-8&12&10\\& & 10& -10& \color{black}{-10} \\ \hline &\color{blue}{-2}&\color{blue}{2}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -2x^{3}-8x^{2}+12x+10 }{ x+5 } = \color{blue}{-2x^{2}+2x+2} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&-2&-8&12&10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ -2 }&-8&12&10\\& & & & \\ \hline &\color{orangered}{-2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&-2&-8&12&10\\& & \color{blue}{10} & & \\ \hline &\color{blue}{-2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 10 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-5&-2&\color{orangered}{ -8 }&12&10\\& & \color{orangered}{10} & & \\ \hline &-2&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&-2&-8&12&10\\& & 10& \color{blue}{-10} & \\ \hline &-2&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-5&-2&-8&\color{orangered}{ 12 }&10\\& & 10& \color{orangered}{-10} & \\ \hline &-2&2&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&-2&-8&12&10\\& & 10& -10& \color{blue}{-10} \\ \hline &-2&2&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-5&-2&-8&12&\color{orangered}{ 10 }\\& & 10& -10& \color{orangered}{-10} \\ \hline &\color{blue}{-2}&\color{blue}{2}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -2x^{2}+2x+2 } $ with a remainder of $ \color{red}{ 0 } $.