Divide using synthetic division $$ ( x^{4}-4x^{3}+13x^{2}-7x+3 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&-4&13&-7&3\\& & 1& -3& 10& \color{black}{3} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{10}&\color{blue}{3}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ x^{4}-4x^{3}+13x^{2}-7x+3 }{ x-1 } = \color{blue}{x^{3}-3x^{2}+10x+3} ~+~ \frac{ \color{red}{ 6 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-4&13&-7&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&-4&13&-7&3\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-4&13&-7&3\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 1 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ -4 }&13&-7&3\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-4&13&-7&3\\& & 1& \color{blue}{-3} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}1&1&-4&\color{orangered}{ 13 }&-7&3\\& & 1& \color{orangered}{-3} & & \\ \hline &1&-3&\color{orangered}{10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 10 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-4&13&-7&3\\& & 1& -3& \color{blue}{10} & \\ \hline &1&-3&\color{blue}{10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 10 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}1&1&-4&13&\color{orangered}{ -7 }&3\\& & 1& -3& \color{orangered}{10} & \\ \hline &1&-3&10&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-4&13&-7&3\\& & 1& -3& 10& \color{blue}{3} \\ \hline &1&-3&10&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 3 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}1&1&-4&13&-7&\color{orangered}{ 3 }\\& & 1& -3& 10& \color{orangered}{3} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{10}&\color{blue}{3}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3x^{2}+10x+3 } $ with a remainder of $ \color{red}{ 6 } $.