Divide using synthetic division $$ ( x^{3}-13x^{2}+50x-50 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&-13&50&-50\\& & 5& -40& \color{black}{50} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-13x^{2}+50x-50 }{ x-5 } = \color{blue}{x^{2}-8x+10} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-13&50&-50\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 1 }&-13&50&-50\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-13&50&-50\\& & \color{blue}{5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 5 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ -13 }&50&-50\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-13&50&-50\\& & 5& \color{blue}{-40} & \\ \hline &1&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 50 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}5&1&-13&\color{orangered}{ 50 }&-50\\& & 5& \color{orangered}{-40} & \\ \hline &1&-8&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 10 } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-13&50&-50\\& & 5& -40& \color{blue}{50} \\ \hline &1&-8&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ 50 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}5&1&-13&50&\color{orangered}{ -50 }\\& & 5& -40& \color{orangered}{50} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-8x+10 } $ with a remainder of $ \color{red}{ 0 } $.