Find remainder when $ 16x^{4}+24x^{3}-17x^{2}-5x-11 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&16&24&-17&-5&-11\\& & -32& 16& 2& \color{black}{6} \\ \hline &\color{blue}{16}&\color{blue}{-8}&\color{blue}{-1}&\color{blue}{-3}&\color{orangered}{-5} \end{array} $$The remainder when $ 16x^{4}+24x^{3}-17x^{2}-5x-11 $ is divided by $ x+2 $ is $ \, \color{red}{ -5 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&16&24&-17&-5&-11\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 16 }&24&-17&-5&-11\\& & & & & \\ \hline &\color{orangered}{16}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 16 } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&16&24&-17&-5&-11\\& & \color{blue}{-32} & & & \\ \hline &\color{blue}{16}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-2&16&\color{orangered}{ 24 }&-17&-5&-11\\& & \color{orangered}{-32} & & & \\ \hline &16&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&16&24&-17&-5&-11\\& & -32& \color{blue}{16} & & \\ \hline &16&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 16 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-2&16&24&\color{orangered}{ -17 }&-5&-11\\& & -32& \color{orangered}{16} & & \\ \hline &16&-8&\color{orangered}{-1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&16&24&-17&-5&-11\\& & -32& 16& \color{blue}{2} & \\ \hline &16&-8&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 2 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-2&16&24&-17&\color{orangered}{ -5 }&-11\\& & -32& 16& \color{orangered}{2} & \\ \hline &16&-8&-1&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&16&24&-17&-5&-11\\& & -32& 16& 2& \color{blue}{6} \\ \hline &16&-8&-1&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 6 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-2&16&24&-17&-5&\color{orangered}{ -11 }\\& & -32& 16& 2& \color{orangered}{6} \\ \hline &\color{blue}{16}&\color{blue}{-8}&\color{blue}{-1}&\color{blue}{-3}&\color{orangered}{-5} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -5 }\right) $.