Find remainder when $ 5x^{2}-12x-40 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}4&5&-12&-40\\& & 20& \color{black}{32} \\ \hline &\color{blue}{5}&\color{blue}{8}&\color{orangered}{-8} \end{array} $$The remainder when $ 5x^{2}-12x-40 $ is divided by $ x-4 $ is $ \, \color{red}{ -8 } $.
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.
$$ \begin{array}{c|rrr}\color{blue}{4}&5&-12&-40\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}4&\color{orangered}{ 5 }&-12&-40\\& & & \\ \hline &\color{orangered}{5}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrr}\color{blue}{4}&5&-12&-40\\& & \color{blue}{20} & \\ \hline &\color{blue}{5}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 20 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrr}4&5&\color{orangered}{ -12 }&-40\\& & \color{orangered}{20} & \\ \hline &5&\color{orangered}{8}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 8 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrr}\color{blue}{4}&5&-12&-40\\& & 20& \color{blue}{32} \\ \hline &5&\color{blue}{8}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 32 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrr}4&5&-12&\color{orangered}{ -40 }\\& & 20& \color{orangered}{32} \\ \hline &\color{blue}{5}&\color{blue}{8}&\color{orangered}{-8} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -8 }\right) $.