Find remainder when $ 2x^{5}+7x^{4}-15x^{3}-x^{2}-16 $ is divided by $ x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-5&2&7&-15&-1&0&-16\\& & -10& 15& 0& 5& \color{black}{-25} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{0}&\color{blue}{-1}&\color{blue}{5}&\color{orangered}{-41} \end{array} $$The remainder when $ 2x^{5}+7x^{4}-15x^{3}-x^{2}-16 $ is divided by $ x+5 $ is $ \, \color{red}{ -41 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&7&-15&-1&0&-16\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-5&\color{orangered}{ 2 }&7&-15&-1&0&-16\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&7&-15&-1&0&-16\\& & \color{blue}{-10} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-5&2&\color{orangered}{ 7 }&-15&-1&0&-16\\& & \color{orangered}{-10} & & & & \\ \hline &2&\color{orangered}{-3}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&7&-15&-1&0&-16\\& & -10& \color{blue}{15} & & & \\ \hline &2&\color{blue}{-3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 15 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-5&2&7&\color{orangered}{ -15 }&-1&0&-16\\& & -10& \color{orangered}{15} & & & \\ \hline &2&-3&\color{orangered}{0}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&7&-15&-1&0&-16\\& & -10& 15& \color{blue}{0} & & \\ \hline &2&-3&\color{blue}{0}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 0 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-5&2&7&-15&\color{orangered}{ -1 }&0&-16\\& & -10& 15& \color{orangered}{0} & & \\ \hline &2&-3&0&\color{orangered}{-1}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&7&-15&-1&0&-16\\& & -10& 15& 0& \color{blue}{5} & \\ \hline &2&-3&0&\color{blue}{-1}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 5 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}-5&2&7&-15&-1&\color{orangered}{ 0 }&-16\\& & -10& 15& 0& \color{orangered}{5} & \\ \hline &2&-3&0&-1&\color{orangered}{5}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 5 } = \color{blue}{ -25 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&7&-15&-1&0&-16\\& & -10& 15& 0& 5& \color{blue}{-25} \\ \hline &2&-3&0&-1&\color{blue}{5}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ \left( -25 \right) } = \color{orangered}{ -41 } $
$$ \begin{array}{c|rrrrrr}-5&2&7&-15&-1&0&\color{orangered}{ -16 }\\& & -10& 15& 0& 5& \color{orangered}{-25} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{0}&\color{blue}{-1}&\color{blue}{5}&\color{orangered}{-41} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -41 }\right) $.