Divide using synthetic division $$ ( 2x^{4}+3x^{3}-7x^{2}-10x ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&2&3&-7&-10&0\\& & 4& 14& 14& \color{black}{8} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{blue}{7}&\color{blue}{4}&\color{orangered}{8} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+3x^{3}-7x^{2}-10x }{ x-2 } = \color{blue}{2x^{3}+7x^{2}+7x+4} ~+~ \frac{ \color{red}{ 8 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&3&-7&-10&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 2 }&3&-7&-10&0\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&3&-7&-10&0\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 4 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}2&2&\color{orangered}{ 3 }&-7&-10&0\\& & \color{orangered}{4} & & & \\ \hline &2&\color{orangered}{7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&3&-7&-10&0\\& & 4& \color{blue}{14} & & \\ \hline &2&\color{blue}{7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 14 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}2&2&3&\color{orangered}{ -7 }&-10&0\\& & 4& \color{orangered}{14} & & \\ \hline &2&7&\color{orangered}{7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&3&-7&-10&0\\& & 4& 14& \color{blue}{14} & \\ \hline &2&7&\color{blue}{7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 14 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}2&2&3&-7&\color{orangered}{ -10 }&0\\& & 4& 14& \color{orangered}{14} & \\ \hline &2&7&7&\color{orangered}{4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&3&-7&-10&0\\& & 4& 14& 14& \color{blue}{8} \\ \hline &2&7&7&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}2&2&3&-7&-10&\color{orangered}{ 0 }\\& & 4& 14& 14& \color{orangered}{8} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{blue}{7}&\color{blue}{4}&\color{orangered}{8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+7x^{2}+7x+4 } $ with a remainder of $ \color{red}{ 8 } $.