Divide using synthetic division $$ ( 6x^{3}+62x^{2}+11x-86 ) \div ( x+10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-10&6&62&11&-86\\& & -60& -20& \color{black}{90} \\ \hline &\color{blue}{6}&\color{blue}{2}&\color{blue}{-9}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+62x^{2}+11x-86 }{ x+10 } = \color{blue}{6x^{2}+2x-9} ~+~ \frac{ \color{red}{ 4 } }{ x+10 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 10 = 0 $ ( $ x = \color{blue}{ -10 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&6&62&11&-86\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-10&\color{orangered}{ 6 }&62&11&-86\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 6 } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&6&62&11&-86\\& & \color{blue}{-60} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 62 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-10&6&\color{orangered}{ 62 }&11&-86\\& & \color{orangered}{-60} & & \\ \hline &6&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 2 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&6&62&11&-86\\& & -60& \color{blue}{-20} & \\ \hline &6&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-10&6&62&\color{orangered}{ 11 }&-86\\& & -60& \color{orangered}{-20} & \\ \hline &6&2&\color{orangered}{-9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 90 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&6&62&11&-86\\& & -60& -20& \color{blue}{90} \\ \hline &6&2&\color{blue}{-9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -86 } + \color{orangered}{ 90 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-10&6&62&11&\color{orangered}{ -86 }\\& & -60& -20& \color{orangered}{90} \\ \hline &\color{blue}{6}&\color{blue}{2}&\color{blue}{-9}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+2x-9 } $ with a remainder of $ \color{red}{ 4 } $.