Divide using synthetic division $$ ( x^{4}-15x^{3}+40x^{2}+52x-5 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&-15&40&52&-5\\& & 5& -50& -50& \color{black}{10} \\ \hline &\color{blue}{1}&\color{blue}{-10}&\color{blue}{-10}&\color{blue}{2}&\color{orangered}{5} \end{array} $$The solution is:
$$ \frac{ x^{4}-15x^{3}+40x^{2}+52x-5 }{ x-5 } = \color{blue}{x^{3}-10x^{2}-10x+2} ~+~ \frac{ \color{red}{ 5 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-15&40&52&-5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&-15&40&52&-5\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-15&40&52&-5\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 5 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ -15 }&40&52&-5\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -50 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-15&40&52&-5\\& & 5& \color{blue}{-50} & & \\ \hline &1&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -50 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}5&1&-15&\color{orangered}{ 40 }&52&-5\\& & 5& \color{orangered}{-50} & & \\ \hline &1&-10&\color{orangered}{-10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -50 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-15&40&52&-5\\& & 5& -50& \color{blue}{-50} & \\ \hline &1&-10&\color{blue}{-10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 52 } + \color{orangered}{ \left( -50 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}5&1&-15&40&\color{orangered}{ 52 }&-5\\& & 5& -50& \color{orangered}{-50} & \\ \hline &1&-10&-10&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-15&40&52&-5\\& & 5& -50& -50& \color{blue}{10} \\ \hline &1&-10&-10&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 10 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}5&1&-15&40&52&\color{orangered}{ -5 }\\& & 5& -50& -50& \color{orangered}{10} \\ \hline &\color{blue}{1}&\color{blue}{-10}&\color{blue}{-10}&\color{blue}{2}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-10x^{2}-10x+2 } $ with a remainder of $ \color{red}{ 5 } $.