Divide using synthetic division $$ ( x^{4}+12x^{3}+25x^{2}+48x-18 ) \div ( x+10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-10&1&12&25&48&-18\\& & -10& -20& -50& \color{black}{20} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{blue}{-2}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ x^{4}+12x^{3}+25x^{2}+48x-18 }{ x+10 } = \color{blue}{x^{3}+2x^{2}+5x-2} ~+~ \frac{ \color{red}{ 2 } }{ x+10 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&1&12&25&48&-18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-10&\color{orangered}{ 1 }&12&25&48&-18\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 1 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&1&12&25&48&-18\\& & \color{blue}{-10} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-10&1&\color{orangered}{ 12 }&25&48&-18\\& & \color{orangered}{-10} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&1&12&25&48&-18\\& & -10& \color{blue}{-20} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-10&1&12&\color{orangered}{ 25 }&48&-18\\& & -10& \color{orangered}{-20} & & \\ \hline &1&2&\color{orangered}{5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 5 } = \color{blue}{ -50 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&1&12&25&48&-18\\& & -10& -20& \color{blue}{-50} & \\ \hline &1&2&\color{blue}{5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 48 } + \color{orangered}{ \left( -50 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-10&1&12&25&\color{orangered}{ 48 }&-18\\& & -10& -20& \color{orangered}{-50} & \\ \hline &1&2&5&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&1&12&25&48&-18\\& & -10& -20& -50& \color{blue}{20} \\ \hline &1&2&5&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 20 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-10&1&12&25&48&\color{orangered}{ -18 }\\& & -10& -20& -50& \color{orangered}{20} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{blue}{-2}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}+5x-2 } $ with a remainder of $ \color{red}{ 2 } $.