Find remainder when $ 11x^{4}+28x^{3}-18x^{2}-4x+13 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&11&28&-18&-4&13\\& & -33& 15& 9& \color{black}{-15} \\ \hline &\color{blue}{11}&\color{blue}{-5}&\color{blue}{-3}&\color{blue}{5}&\color{orangered}{-2} \end{array} $$The remainder when $ 11x^{4}+28x^{3}-18x^{2}-4x+13 $ is divided by $ x+3 $ is $ \, \color{red}{ -2 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&11&28&-18&-4&13\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 11 }&28&-18&-4&13\\& & & & & \\ \hline &\color{orangered}{11}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 11 } = \color{blue}{ -33 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&11&28&-18&-4&13\\& & \color{blue}{-33} & & & \\ \hline &\color{blue}{11}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -33 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-3&11&\color{orangered}{ 28 }&-18&-4&13\\& & \color{orangered}{-33} & & & \\ \hline &11&\color{orangered}{-5}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&11&28&-18&-4&13\\& & -33& \color{blue}{15} & & \\ \hline &11&\color{blue}{-5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 15 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-3&11&28&\color{orangered}{ -18 }&-4&13\\& & -33& \color{orangered}{15} & & \\ \hline &11&-5&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&11&28&-18&-4&13\\& & -33& 15& \color{blue}{9} & \\ \hline &11&-5&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 9 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-3&11&28&-18&\color{orangered}{ -4 }&13\\& & -33& 15& \color{orangered}{9} & \\ \hline &11&-5&-3&\color{orangered}{5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&11&28&-18&-4&13\\& & -33& 15& 9& \color{blue}{-15} \\ \hline &11&-5&-3&\color{blue}{5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-3&11&28&-18&-4&\color{orangered}{ 13 }\\& & -33& 15& 9& \color{orangered}{-15} \\ \hline &\color{blue}{11}&\color{blue}{-5}&\color{blue}{-3}&\color{blue}{5}&\color{orangered}{-2} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -2 }\right) $.