Divide using synthetic division $$ ( x^{3}-7x^{2}+13x+3 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&-7&13&3\\& & 2& -10& \color{black}{6} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{3}&\color{orangered}{9} \end{array} $$The solution is:
$$ \frac{ x^{3}-7x^{2}+13x+3 }{ x-2 } = \color{blue}{x^{2}-5x+3} ~+~ \frac{ \color{red}{ 9 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-7&13&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&-7&13&3\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-7&13&3\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 2 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ -7 }&13&3\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-7&13&3\\& & 2& \color{blue}{-10} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}2&1&-7&\color{orangered}{ 13 }&3\\& & 2& \color{orangered}{-10} & \\ \hline &1&-5&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-7&13&3\\& & 2& -10& \color{blue}{6} \\ \hline &1&-5&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 6 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}2&1&-7&13&\color{orangered}{ 3 }\\& & 2& -10& \color{orangered}{6} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{3}&\color{orangered}{9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-5x+3 } $ with a remainder of $ \color{red}{ 9 } $.