Find remainder when $ 2x^{2}+20x-59 $ is divided by $ x-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}5&2&20&-59\\& & 10& \color{black}{150} \\ \hline &\color{blue}{2}&\color{blue}{30}&\color{orangered}{91} \end{array} $$The remainder when $ 2x^{2}+20x-59 $ is divided by $ x-5 $ is $ \, \color{red}{ 91 } $.
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 -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{5}&2&20&-59\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}5&\color{orangered}{ 2 }&20&-59\\& & & \\ \hline &\color{orangered}{2}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&2&20&-59\\& & \color{blue}{10} & \\ \hline &\color{blue}{2}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ 10 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrr}5&2&\color{orangered}{ 20 }&-59\\& & \color{orangered}{10} & \\ \hline &2&\color{orangered}{30}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 30 } = \color{blue}{ 150 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&2&20&-59\\& & 10& \color{blue}{150} \\ \hline &2&\color{blue}{30}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -59 } + \color{orangered}{ 150 } = \color{orangered}{ 91 } $
$$ \begin{array}{c|rrr}5&2&20&\color{orangered}{ -59 }\\& & 10& \color{orangered}{150} \\ \hline &\color{blue}{2}&\color{blue}{30}&\color{orangered}{91} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 91 }\right) $.