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