Find remainder when $ 6x^{2}+25x+31 $ is divided by $ x-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}5&6&25&31\\& & 30& \color{black}{275} \\ \hline &\color{blue}{6}&\color{blue}{55}&\color{orangered}{306} \end{array} $$The remainder when $ 6x^{2}+25x+31 $ is divided by $ x-5 $ is $ \, \color{red}{ 306 } $.
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.
$$ \begin{array}{c|rrr}\color{blue}{5}&6&25&31\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}5&\color{orangered}{ 6 }&25&31\\& & & \\ \hline &\color{orangered}{6}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 6 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&6&25&31\\& & \color{blue}{30} & \\ \hline &\color{blue}{6}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ 30 } = \color{orangered}{ 55 } $
$$ \begin{array}{c|rrr}5&6&\color{orangered}{ 25 }&31\\& & \color{orangered}{30} & \\ \hline &6&\color{orangered}{55}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 55 } = \color{blue}{ 275 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&6&25&31\\& & 30& \color{blue}{275} \\ \hline &6&\color{blue}{55}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 31 } + \color{orangered}{ 275 } = \color{orangered}{ 306 } $
$$ \begin{array}{c|rrr}5&6&25&\color{orangered}{ 31 }\\& & 30& \color{orangered}{275} \\ \hline &\color{blue}{6}&\color{blue}{55}&\color{orangered}{306} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 306 }\right) $.