Find remainder when $ 2x^{4}-6x^{3}-23x^{2}-5x+5 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&2&-6&-23&-5&5\\& & -4& 20& 6& \color{black}{-2} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{-3}&\color{blue}{1}&\color{orangered}{3} \end{array} $$The remainder when $ 2x^{4}-6x^{3}-23x^{2}-5x+5 $ is divided by $ x+2 $ is $ \, \color{red}{ 3 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-6&-23&-5&5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 2 }&-6&-23&-5&5\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-6&-23&-5&5\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-2&2&\color{orangered}{ -6 }&-23&-5&5\\& & \color{orangered}{-4} & & & \\ \hline &2&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-6&-23&-5&5\\& & -4& \color{blue}{20} & & \\ \hline &2&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 20 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-2&2&-6&\color{orangered}{ -23 }&-5&5\\& & -4& \color{orangered}{20} & & \\ \hline &2&-10&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-6&-23&-5&5\\& & -4& 20& \color{blue}{6} & \\ \hline &2&-10&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 6 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-2&2&-6&-23&\color{orangered}{ -5 }&5\\& & -4& 20& \color{orangered}{6} & \\ \hline &2&-10&-3&\color{orangered}{1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-6&-23&-5&5\\& & -4& 20& 6& \color{blue}{-2} \\ \hline &2&-10&-3&\color{blue}{1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-2&2&-6&-23&-5&\color{orangered}{ 5 }\\& & -4& 20& 6& \color{orangered}{-2} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{-3}&\color{blue}{1}&\color{orangered}{3} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 3 }\right) $.