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