Find remainder when $ 15x^{4}-2x^{3}-6x+36 $ is divided by $ x+1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&15&-2&0&-6&36\\& & -15& 17& -17& \color{black}{23} \\ \hline &\color{blue}{15}&\color{blue}{-17}&\color{blue}{17}&\color{blue}{-23}&\color{orangered}{59} \end{array} $$The remainder when $ 15x^{4}-2x^{3}-6x+36 $ is divided by $ x+1 $ is $ \, \color{red}{ 59 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&15&-2&0&-6&36\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 15 }&-2&0&-6&36\\& & & & & \\ \hline &\color{orangered}{15}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 15 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&15&-2&0&-6&36\\& & \color{blue}{-15} & & & \\ \hline &\color{blue}{15}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrrr}-1&15&\color{orangered}{ -2 }&0&-6&36\\& & \color{orangered}{-15} & & & \\ \hline &15&\color{orangered}{-17}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -17 \right) } = \color{blue}{ 17 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&15&-2&0&-6&36\\& & -15& \color{blue}{17} & & \\ \hline &15&\color{blue}{-17}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 17 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}-1&15&-2&\color{orangered}{ 0 }&-6&36\\& & -15& \color{orangered}{17} & & \\ \hline &15&-17&\color{orangered}{17}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 17 } = \color{blue}{ -17 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&15&-2&0&-6&36\\& & -15& 17& \color{blue}{-17} & \\ \hline &15&-17&\color{blue}{17}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -17 \right) } = \color{orangered}{ -23 } $
$$ \begin{array}{c|rrrrr}-1&15&-2&0&\color{orangered}{ -6 }&36\\& & -15& 17& \color{orangered}{-17} & \\ \hline &15&-17&17&\color{orangered}{-23}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -23 \right) } = \color{blue}{ 23 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&15&-2&0&-6&36\\& & -15& 17& -17& \color{blue}{23} \\ \hline &15&-17&17&\color{blue}{-23}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 36 } + \color{orangered}{ 23 } = \color{orangered}{ 59 } $
$$ \begin{array}{c|rrrrr}-1&15&-2&0&-6&\color{orangered}{ 36 }\\& & -15& 17& -17& \color{orangered}{23} \\ \hline &\color{blue}{15}&\color{blue}{-17}&\color{blue}{17}&\color{blue}{-23}&\color{orangered}{59} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 59 }\right) $.