Divide using synthetic division $$ ( x^{3}-13x^{2}+25x+50 ) \div ( x-10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}10&1&-13&25&50\\& & 10& -30& \color{black}{-50} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-13x^{2}+25x+50 }{ x-10 } = \color{blue}{x^{2}-3x-5} $$Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{10}&1&-13&25&50\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}10&\color{orangered}{ 1 }&-13&25&50\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 1 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&1&-13&25&50\\& & \color{blue}{10} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 10 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}10&1&\color{orangered}{ -13 }&25&50\\& & \color{orangered}{10} & & \\ \hline &1&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&1&-13&25&50\\& & 10& \color{blue}{-30} & \\ \hline &1&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}10&1&-13&\color{orangered}{ 25 }&50\\& & 10& \color{orangered}{-30} & \\ \hline &1&-3&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&1&-13&25&50\\& & 10& -30& \color{blue}{-50} \\ \hline &1&-3&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 50 } + \color{orangered}{ \left( -50 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}10&1&-13&25&\color{orangered}{ 50 }\\& & 10& -30& \color{orangered}{-50} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-3x-5 } $ with a remainder of $ \color{red}{ 0 } $.