Find remainder when $ 10x^{4}+22x^{3}+27x^{2}+3x+15 $ is divided by $ x+1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&10&22&27&3&15\\& & -10& -12& -15& \color{black}{12} \\ \hline &\color{blue}{10}&\color{blue}{12}&\color{blue}{15}&\color{blue}{-12}&\color{orangered}{27} \end{array} $$The remainder when $ 10x^{4}+22x^{3}+27x^{2}+3x+15 $ is divided by $ x+1 $ 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 + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&10&22&27&3&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 10 }&22&27&3&15\\& & & & & \\ \hline &\color{orangered}{10}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 10 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&10&22&27&3&15\\& & \color{blue}{-10} & & & \\ \hline &\color{blue}{10}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-1&10&\color{orangered}{ 22 }&27&3&15\\& & \color{orangered}{-10} & & & \\ \hline &10&\color{orangered}{12}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 12 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&10&22&27&3&15\\& & -10& \color{blue}{-12} & & \\ \hline &10&\color{blue}{12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}-1&10&22&\color{orangered}{ 27 }&3&15\\& & -10& \color{orangered}{-12} & & \\ \hline &10&12&\color{orangered}{15}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 15 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&10&22&27&3&15\\& & -10& -12& \color{blue}{-15} & \\ \hline &10&12&\color{blue}{15}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}-1&10&22&27&\color{orangered}{ 3 }&15\\& & -10& -12& \color{orangered}{-15} & \\ \hline &10&12&15&\color{orangered}{-12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&10&22&27&3&15\\& & -10& -12& -15& \color{blue}{12} \\ \hline &10&12&15&\color{blue}{-12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 12 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}-1&10&22&27&3&\color{orangered}{ 15 }\\& & -10& -12& -15& \color{orangered}{12} \\ \hline &\color{blue}{10}&\color{blue}{12}&\color{blue}{15}&\color{blue}{-12}&\color{orangered}{27} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 27 }\right) $.