Find remainder when $ x^{4}+10x^{3}-24 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&10&0&0&-24\\& & -2& -16& 32& \color{black}{-64} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{-16}&\color{blue}{32}&\color{orangered}{-88} \end{array} $$The remainder when $ x^{4}+10x^{3}-24 $ is divided by $ x+2 $ is $ \, \color{red}{ -88 } $.
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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&10&0&0&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&10&0&0&-24\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&10&0&0&-24\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ 10 }&0&0&-24\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&10&0&0&-24\\& & -2& \color{blue}{-16} & & \\ \hline &1&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}-2&1&10&\color{orangered}{ 0 }&0&-24\\& & -2& \color{orangered}{-16} & & \\ \hline &1&8&\color{orangered}{-16}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&10&0&0&-24\\& & -2& -16& \color{blue}{32} & \\ \hline &1&8&\color{blue}{-16}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 32 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrr}-2&1&10&0&\color{orangered}{ 0 }&-24\\& & -2& -16& \color{orangered}{32} & \\ \hline &1&8&-16&\color{orangered}{32}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 32 } = \color{blue}{ -64 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&10&0&0&-24\\& & -2& -16& 32& \color{blue}{-64} \\ \hline &1&8&-16&\color{blue}{32}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ \left( -64 \right) } = \color{orangered}{ -88 } $
$$ \begin{array}{c|rrrrr}-2&1&10&0&0&\color{orangered}{ -24 }\\& & -2& -16& 32& \color{orangered}{-64} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{-16}&\color{blue}{32}&\color{orangered}{-88} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -88 }\right) $.