Divide using synthetic division $$ ( 2x^{5}-2x^{3}+4x^{2}-3 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}1&2&0&-2&4&0&-3\\& & 2& 2& 0& 4& \color{black}{4} \\ \hline &\color{blue}{2}&\color{blue}{2}&\color{blue}{0}&\color{blue}{4}&\color{blue}{4}&\color{orangered}{1} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-2x^{3}+4x^{2}-3 }{ x-1 } = \color{blue}{2x^{4}+2x^{3}+4x+4} ~+~ \frac{ \color{red}{ 1 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&0&-2&4&0&-3\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}1&\color{orangered}{ 2 }&0&-2&4&0&-3\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&0&-2&4&0&-3\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}1&2&\color{orangered}{ 0 }&-2&4&0&-3\\& & \color{orangered}{2} & & & & \\ \hline &2&\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&0&-2&4&0&-3\\& & 2& \color{blue}{2} & & & \\ \hline &2&\color{blue}{2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 2 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}1&2&0&\color{orangered}{ -2 }&4&0&-3\\& & 2& \color{orangered}{2} & & & \\ \hline &2&2&\color{orangered}{0}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&0&-2&4&0&-3\\& & 2& 2& \color{blue}{0} & & \\ \hline &2&2&\color{blue}{0}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 0 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}1&2&0&-2&\color{orangered}{ 4 }&0&-3\\& & 2& 2& \color{orangered}{0} & & \\ \hline &2&2&0&\color{orangered}{4}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&0&-2&4&0&-3\\& & 2& 2& 0& \color{blue}{4} & \\ \hline &2&2&0&\color{blue}{4}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}1&2&0&-2&4&\color{orangered}{ 0 }&-3\\& & 2& 2& 0& \color{orangered}{4} & \\ \hline &2&2&0&4&\color{orangered}{4}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&0&-2&4&0&-3\\& & 2& 2& 0& 4& \color{blue}{4} \\ \hline &2&2&0&4&\color{blue}{4}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 4 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}1&2&0&-2&4&0&\color{orangered}{ -3 }\\& & 2& 2& 0& 4& \color{orangered}{4} \\ \hline &\color{blue}{2}&\color{blue}{2}&\color{blue}{0}&\color{blue}{4}&\color{blue}{4}&\color{orangered}{1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}+2x^{3}+4x+4 } $ with a remainder of $ \color{red}{ 1 } $.