Divide using synthetic division $$ ( 2x^{4}-10x^{3}+x-7 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&2&-10&0&1&-7\\& & 10& 0& 0& \color{black}{5} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{1}&\color{orangered}{-2} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-10x^{3}+x-7 }{ x-5 } = \color{blue}{2x^{3}+1} \color{red}{~-~} \frac{ \color{red}{ 2 } }{ x-5 } $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-10&0&1&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 2 }&-10&0&1&-7\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-10&0&1&-7\\& & \color{blue}{10} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 10 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&2&\color{orangered}{ -10 }&0&1&-7\\& & \color{orangered}{10} & & & \\ \hline &2&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-10&0&1&-7\\& & 10& \color{blue}{0} & & \\ \hline &2&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&2&-10&\color{orangered}{ 0 }&1&-7\\& & 10& \color{orangered}{0} & & \\ \hline &2&0&\color{orangered}{0}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-10&0&1&-7\\& & 10& 0& \color{blue}{0} & \\ \hline &2&0&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 0 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}5&2&-10&0&\color{orangered}{ 1 }&-7\\& & 10& 0& \color{orangered}{0} & \\ \hline &2&0&0&\color{orangered}{1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-10&0&1&-7\\& & 10& 0& 0& \color{blue}{5} \\ \hline &2&0&0&\color{blue}{1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 5 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}5&2&-10&0&1&\color{orangered}{ -7 }\\& & 10& 0& 0& \color{orangered}{5} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{1}&\color{orangered}{-2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+1 } $ with a remainder of $ \color{red}{ -2 } $.