Find remainder when $ 2x^{3}+x^{2}-6x-8 $ is divided by $ x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&2&1&-6&-8\\& & 4& 10& \color{black}{8} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The remainder when $ 2x^{3}+x^{2}-6x-8 $ is divided by $ x-2 $ is $ \, \color{red}{ 0 } $.
Explanation
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&1&-6&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 2 }&1&-6&-8\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&1&-6&-8\\& & \color{blue}{4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}2&2&\color{orangered}{ 1 }&-6&-8\\& & \color{orangered}{4} & & \\ \hline &2&\color{orangered}{5}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&1&-6&-8\\& & 4& \color{blue}{10} & \\ \hline &2&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 10 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}2&2&1&\color{orangered}{ -6 }&-8\\& & 4& \color{orangered}{10} & \\ \hline &2&5&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&1&-6&-8\\& & 4& 10& \color{blue}{8} \\ \hline &2&5&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 8 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&2&1&-6&\color{orangered}{ -8 }\\& & 4& 10& \color{orangered}{8} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.