Divide using synthetic division $$ ( 2x^{5}-3x^{3}-5x^{2}+3x+8 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&2&0&-3&-5&3&8\\& & 4& 8& 10& 10& \color{black}{26} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{5}&\color{blue}{5}&\color{blue}{13}&\color{orangered}{34} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-3x^{3}-5x^{2}+3x+8 }{ x-2 } = \color{blue}{2x^{4}+4x^{3}+5x^{2}+5x+13} ~+~ \frac{ \color{red}{ 34 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&0&-3&-5&3&8\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 2 }&0&-3&-5&3&8\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&0&-3&-5&3&8\\& & \color{blue}{4} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}2&2&\color{orangered}{ 0 }&-3&-5&3&8\\& & \color{orangered}{4} & & & & \\ \hline &2&\color{orangered}{4}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&0&-3&-5&3&8\\& & 4& \color{blue}{8} & & & \\ \hline &2&\color{blue}{4}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 8 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}2&2&0&\color{orangered}{ -3 }&-5&3&8\\& & 4& \color{orangered}{8} & & & \\ \hline &2&4&\color{orangered}{5}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&0&-3&-5&3&8\\& & 4& 8& \color{blue}{10} & & \\ \hline &2&4&\color{blue}{5}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 10 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}2&2&0&-3&\color{orangered}{ -5 }&3&8\\& & 4& 8& \color{orangered}{10} & & \\ \hline &2&4&5&\color{orangered}{5}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&0&-3&-5&3&8\\& & 4& 8& 10& \color{blue}{10} & \\ \hline &2&4&5&\color{blue}{5}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 10 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrrr}2&2&0&-3&-5&\color{orangered}{ 3 }&8\\& & 4& 8& 10& \color{orangered}{10} & \\ \hline &2&4&5&5&\color{orangered}{13}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 13 } = \color{blue}{ 26 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&0&-3&-5&3&8\\& & 4& 8& 10& 10& \color{blue}{26} \\ \hline &2&4&5&5&\color{blue}{13}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 26 } = \color{orangered}{ 34 } $
$$ \begin{array}{c|rrrrrr}2&2&0&-3&-5&3&\color{orangered}{ 8 }\\& & 4& 8& 10& 10& \color{orangered}{26} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{5}&\color{blue}{5}&\color{blue}{13}&\color{orangered}{34} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}+4x^{3}+5x^{2}+5x+13 } $ with a remainder of $ \color{red}{ 34 } $.