Find remainder when $ x^{4}+12x^{3}-52x^{2}+48x-18 $ is divided by $ x+10 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-10&1&12&-52&48&-18\\& & -10& -20& 720& \color{black}{-7680} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-72}&\color{blue}{768}&\color{orangered}{-7698} \end{array} $$The remainder when $ x^{4}+12x^{3}-52x^{2}+48x-18 $ is divided by $ x+10 $ is $ \, \color{red}{ -7698 } $.
Explanation
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&1&12&-52&48&-18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-10&\color{orangered}{ 1 }&12&-52&48&-18\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&1&12&-52&48&-18\\& & \color{blue}{-10} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-10&1&\color{orangered}{ 12 }&-52&48&-18\\& & \color{orangered}{-10} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&1&12&-52&48&-18\\& & -10& \color{blue}{-20} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -52 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -72 } $
$$ \begin{array}{c|rrrrr}-10&1&12&\color{orangered}{ -52 }&48&-18\\& & -10& \color{orangered}{-20} & & \\ \hline &1&2&\color{orangered}{-72}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ \left( -72 \right) } = \color{blue}{ 720 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&1&12&-52&48&-18\\& & -10& -20& \color{blue}{720} & \\ \hline &1&2&\color{blue}{-72}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 48 } + \color{orangered}{ 720 } = \color{orangered}{ 768 } $
$$ \begin{array}{c|rrrrr}-10&1&12&-52&\color{orangered}{ 48 }&-18\\& & -10& -20& \color{orangered}{720} & \\ \hline &1&2&-72&\color{orangered}{768}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 768 } = \color{blue}{ -7680 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&1&12&-52&48&-18\\& & -10& -20& 720& \color{blue}{-7680} \\ \hline &1&2&-72&\color{blue}{768}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ \left( -7680 \right) } = \color{orangered}{ -7698 } $
$$ \begin{array}{c|rrrrr}-10&1&12&-52&48&\color{orangered}{ -18 }\\& & -10& -20& 720& \color{orangered}{-7680} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-72}&\color{blue}{768}&\color{orangered}{-7698} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -7698 }\right) $.