Find remainder when $ 4x^{4}+25x^{3}+9x^{2}+16x-12 $ is divided by $ x+6 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-6&4&25&9&16&-12\\& & -24& -6& -18& \color{black}{12} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{3}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The remainder when $ 4x^{4}+25x^{3}+9x^{2}+16x-12 $ is divided by $ x+6 $ is $ \, \color{red}{ 0 } $.
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 + 6 = 0 $ ( $ x = \color{blue}{ -6 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-6}&4&25&9&16&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-6&\color{orangered}{ 4 }&25&9&16&-12\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 4 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-6}&4&25&9&16&-12\\& & \color{blue}{-24} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-6&4&\color{orangered}{ 25 }&9&16&-12\\& & \color{orangered}{-24} & & & \\ \hline &4&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 1 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-6}&4&25&9&16&-12\\& & -24& \color{blue}{-6} & & \\ \hline &4&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-6&4&25&\color{orangered}{ 9 }&16&-12\\& & -24& \color{orangered}{-6} & & \\ \hline &4&1&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 3 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-6}&4&25&9&16&-12\\& & -24& -6& \color{blue}{-18} & \\ \hline &4&1&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-6&4&25&9&\color{orangered}{ 16 }&-12\\& & -24& -6& \color{orangered}{-18} & \\ \hline &4&1&3&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-6}&4&25&9&16&-12\\& & -24& -6& -18& \color{blue}{12} \\ \hline &4&1&3&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-6&4&25&9&16&\color{orangered}{ -12 }\\& & -24& -6& -18& \color{orangered}{12} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{3}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.