Find remainder when $ 2x^{3}+5x^{2}-2x+2 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&2&5&-2&2\\& & 8& 52& \color{black}{200} \\ \hline &\color{blue}{2}&\color{blue}{13}&\color{blue}{50}&\color{orangered}{202} \end{array} $$The remainder when $ 2x^{3}+5x^{2}-2x+2 $ is divided by $ x-4 $ is $ \, \color{red}{ 202 } $.
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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&5&-2&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 2 }&5&-2&2\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&5&-2&2\\& & \color{blue}{8} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 8 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}4&2&\color{orangered}{ 5 }&-2&2\\& & \color{orangered}{8} & & \\ \hline &2&\color{orangered}{13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 13 } = \color{blue}{ 52 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&5&-2&2\\& & 8& \color{blue}{52} & \\ \hline &2&\color{blue}{13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 52 } = \color{orangered}{ 50 } $
$$ \begin{array}{c|rrrr}4&2&5&\color{orangered}{ -2 }&2\\& & 8& \color{orangered}{52} & \\ \hline &2&13&\color{orangered}{50}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 50 } = \color{blue}{ 200 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&5&-2&2\\& & 8& 52& \color{blue}{200} \\ \hline &2&13&\color{blue}{50}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 200 } = \color{orangered}{ 202 } $
$$ \begin{array}{c|rrrr}4&2&5&-2&\color{orangered}{ 2 }\\& & 8& 52& \color{orangered}{200} \\ \hline &\color{blue}{2}&\color{blue}{13}&\color{blue}{50}&\color{orangered}{202} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 202 }\right) $.