Find remainder when $ x^{4}-3x^{3}-10x^{2}+3x-7 $ is divided by $ x-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&-3&-10&3&-7\\& & 5& 10& 0& \color{black}{15} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{0}&\color{blue}{3}&\color{orangered}{8} \end{array} $$The remainder when $ x^{4}-3x^{3}-10x^{2}+3x-7 $ is divided by $ x-5 $ is $ \, \color{red}{ 8 } $.
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 -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-3&-10&3&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&-3&-10&3&-7\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-3&-10&3&-7\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 5 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ -3 }&-10&3&-7\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-3&-10&3&-7\\& & 5& \color{blue}{10} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 10 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&1&-3&\color{orangered}{ -10 }&3&-7\\& & 5& \color{orangered}{10} & & \\ \hline &1&2&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-3&-10&3&-7\\& & 5& 10& \color{blue}{0} & \\ \hline &1&2&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 0 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}5&1&-3&-10&\color{orangered}{ 3 }&-7\\& & 5& 10& \color{orangered}{0} & \\ \hline &1&2&0&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-3&-10&3&-7\\& & 5& 10& 0& \color{blue}{15} \\ \hline &1&2&0&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 15 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}5&1&-3&-10&3&\color{orangered}{ -7 }\\& & 5& 10& 0& \color{orangered}{15} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{0}&\color{blue}{3}&\color{orangered}{8} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 8 }\right) $.