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