Find remainder when $ x^{3}-6x^{2}+5 $ is divided by $ x+10 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-10&1&-6&0&5\\& & -10& 160& \color{black}{-1600} \\ \hline &\color{blue}{1}&\color{blue}{-16}&\color{blue}{160}&\color{orangered}{-1595} \end{array} $$The remainder when $ x^{3}-6x^{2}+5 $ is divided by $ x+10 $ is $ \, \color{red}{ -1595 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&-6&0&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-10&\color{orangered}{ 1 }&-6&0&5\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&-6&0&5\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}-10&1&\color{orangered}{ -6 }&0&5\\& & \color{orangered}{-10} & & \\ \hline &1&\color{orangered}{-16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 160 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&-6&0&5\\& & -10& \color{blue}{160} & \\ \hline &1&\color{blue}{-16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 160 } = \color{orangered}{ 160 } $
$$ \begin{array}{c|rrrr}-10&1&-6&\color{orangered}{ 0 }&5\\& & -10& \color{orangered}{160} & \\ \hline &1&-16&\color{orangered}{160}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 160 } = \color{blue}{ -1600 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&-6&0&5\\& & -10& 160& \color{blue}{-1600} \\ \hline &1&-16&\color{blue}{160}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -1600 \right) } = \color{orangered}{ -1595 } $
$$ \begin{array}{c|rrrr}-10&1&-6&0&\color{orangered}{ 5 }\\& & -10& 160& \color{orangered}{-1600} \\ \hline &\color{blue}{1}&\color{blue}{-16}&\color{blue}{160}&\color{orangered}{-1595} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -1595 }\right) $.