Find remainder when $ 4x^{4}+17x^{3}-12x^{2}+10x+26 $ is divided by $ x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&4&17&-12&10&26\\& & -20& 15& -15& \color{black}{25} \\ \hline &\color{blue}{4}&\color{blue}{-3}&\color{blue}{3}&\color{blue}{-5}&\color{orangered}{51} \end{array} $$The remainder when $ 4x^{4}+17x^{3}-12x^{2}+10x+26 $ is divided by $ x+5 $ is $ \, \color{red}{ 51 } $.
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 + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&4&17&-12&10&26\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 4 }&17&-12&10&26\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&4&17&-12&10&26\\& & \color{blue}{-20} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-5&4&\color{orangered}{ 17 }&-12&10&26\\& & \color{orangered}{-20} & & & \\ \hline &4&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&4&17&-12&10&26\\& & -20& \color{blue}{15} & & \\ \hline &4&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 15 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-5&4&17&\color{orangered}{ -12 }&10&26\\& & -20& \color{orangered}{15} & & \\ \hline &4&-3&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 3 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&4&17&-12&10&26\\& & -20& 15& \color{blue}{-15} & \\ \hline &4&-3&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-5&4&17&-12&\color{orangered}{ 10 }&26\\& & -20& 15& \color{orangered}{-15} & \\ \hline &4&-3&3&\color{orangered}{-5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&4&17&-12&10&26\\& & -20& 15& -15& \color{blue}{25} \\ \hline &4&-3&3&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ 25 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrrr}-5&4&17&-12&10&\color{orangered}{ 26 }\\& & -20& 15& -15& \color{orangered}{25} \\ \hline &\color{blue}{4}&\color{blue}{-3}&\color{blue}{3}&\color{blue}{-5}&\color{orangered}{51} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 51 }\right) $.