Divide using synthetic division $$ ( 2x^{4}+x^{3}-3 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&2&1&0&0&-3\\& & 4& 10& 20& \color{black}{40} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{10}&\color{blue}{20}&\color{orangered}{37} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+x^{3}-3 }{ x-2 } = \color{blue}{2x^{3}+5x^{2}+10x+20} ~+~ \frac{ \color{red}{ 37 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&1&0&0&-3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 2 }&1&0&0&-3\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&1&0&0&-3\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}2&2&\color{orangered}{ 1 }&0&0&-3\\& & \color{orangered}{4} & & & \\ \hline &2&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&1&0&0&-3\\& & 4& \color{blue}{10} & & \\ \hline &2&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}2&2&1&\color{orangered}{ 0 }&0&-3\\& & 4& \color{orangered}{10} & & \\ \hline &2&5&\color{orangered}{10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&1&0&0&-3\\& & 4& 10& \color{blue}{20} & \\ \hline &2&5&\color{blue}{10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 20 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}2&2&1&0&\color{orangered}{ 0 }&-3\\& & 4& 10& \color{orangered}{20} & \\ \hline &2&5&10&\color{orangered}{20}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 20 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&1&0&0&-3\\& & 4& 10& 20& \color{blue}{40} \\ \hline &2&5&10&\color{blue}{20}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 40 } = \color{orangered}{ 37 } $
$$ \begin{array}{c|rrrrr}2&2&1&0&0&\color{orangered}{ -3 }\\& & 4& 10& 20& \color{orangered}{40} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{10}&\color{blue}{20}&\color{orangered}{37} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+5x^{2}+10x+20 } $ with a remainder of $ \color{red}{ 37 } $.