Find remainder when $ 6x^{2}+33x-12 $ is divided by $ x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}-5&6&33&-12\\& & -30& \color{black}{-15} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{orangered}{-27} \end{array} $$The remainder when $ 6x^{2}+33x-12 $ is divided by $ x+5 $ is $ \, \color{red}{ -27 } $.
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.
$$ \begin{array}{c|rrr}\color{blue}{-5}&6&33&-12\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}-5&\color{orangered}{ 6 }&33&-12\\& & & \\ \hline &\color{orangered}{6}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{-5}&6&33&-12\\& & \color{blue}{-30} & \\ \hline &\color{blue}{6}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 33 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrr}-5&6&\color{orangered}{ 33 }&-12\\& & \color{orangered}{-30} & \\ \hline &6&\color{orangered}{3}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 3 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrr}\color{blue}{-5}&6&33&-12\\& & -30& \color{blue}{-15} \\ \hline &6&\color{blue}{3}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -27 } $
$$ \begin{array}{c|rrr}-5&6&33&\color{orangered}{ -12 }\\& & -30& \color{orangered}{-15} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{orangered}{-27} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -27 }\right) $.