Find remainder when $ -2x^{4}+x^{3}-8x^{2}+10 $ is divided by $ x+6 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-6&-2&1&-8&0&10\\& & 12& -78& 516& \color{black}{-3096} \\ \hline &\color{blue}{-2}&\color{blue}{13}&\color{blue}{-86}&\color{blue}{516}&\color{orangered}{-3086} \end{array} $$The remainder when $ -2x^{4}+x^{3}-8x^{2}+10 $ is divided by $ x+6 $ is $ \, \color{red}{ -3086 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-6}&-2&1&-8&0&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-6&\color{orangered}{ -2 }&1&-8&0&10\\& & & & & \\ \hline &\color{orangered}{-2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-6}&-2&1&-8&0&10\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{-2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 12 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}-6&-2&\color{orangered}{ 1 }&-8&0&10\\& & \color{orangered}{12} & & & \\ \hline &-2&\color{orangered}{13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 13 } = \color{blue}{ -78 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-6}&-2&1&-8&0&10\\& & 12& \color{blue}{-78} & & \\ \hline &-2&\color{blue}{13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -78 \right) } = \color{orangered}{ -86 } $
$$ \begin{array}{c|rrrrr}-6&-2&1&\color{orangered}{ -8 }&0&10\\& & 12& \color{orangered}{-78} & & \\ \hline &-2&13&\color{orangered}{-86}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -86 \right) } = \color{blue}{ 516 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-6}&-2&1&-8&0&10\\& & 12& -78& \color{blue}{516} & \\ \hline &-2&13&\color{blue}{-86}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 516 } = \color{orangered}{ 516 } $
$$ \begin{array}{c|rrrrr}-6&-2&1&-8&\color{orangered}{ 0 }&10\\& & 12& -78& \color{orangered}{516} & \\ \hline &-2&13&-86&\color{orangered}{516}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 516 } = \color{blue}{ -3096 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-6}&-2&1&-8&0&10\\& & 12& -78& 516& \color{blue}{-3096} \\ \hline &-2&13&-86&\color{blue}{516}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -3096 \right) } = \color{orangered}{ -3086 } $
$$ \begin{array}{c|rrrrr}-6&-2&1&-8&0&\color{orangered}{ 10 }\\& & 12& -78& 516& \color{orangered}{-3096} \\ \hline &\color{blue}{-2}&\color{blue}{13}&\color{blue}{-86}&\color{blue}{516}&\color{orangered}{-3086} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -3086 }\right) $.