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