Find remainder when $ 3x^{4}+13x^{3}-4x+15 $ is divided by $ x+4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&3&13&0&-4&15\\& & -12& -4& 16& \color{black}{-48} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{-4}&\color{blue}{12}&\color{orangered}{-33} \end{array} $$The remainder when $ 3x^{4}+13x^{3}-4x+15 $ is divided by $ x+4 $ is $ \, \color{red}{ -33 } $.
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 + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&13&0&-4&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 3 }&13&0&-4&15\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 3 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&13&0&-4&15\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-4&3&\color{orangered}{ 13 }&0&-4&15\\& & \color{orangered}{-12} & & & \\ \hline &3&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&13&0&-4&15\\& & -12& \color{blue}{-4} & & \\ \hline &3&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-4&3&13&\color{orangered}{ 0 }&-4&15\\& & -12& \color{orangered}{-4} & & \\ \hline &3&1&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&13&0&-4&15\\& & -12& -4& \color{blue}{16} & \\ \hline &3&1&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 16 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-4&3&13&0&\color{orangered}{ -4 }&15\\& & -12& -4& \color{orangered}{16} & \\ \hline &3&1&-4&\color{orangered}{12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 12 } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&13&0&-4&15\\& & -12& -4& 16& \color{blue}{-48} \\ \hline &3&1&-4&\color{blue}{12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -33 } $
$$ \begin{array}{c|rrrrr}-4&3&13&0&-4&\color{orangered}{ 15 }\\& & -12& -4& 16& \color{orangered}{-48} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{-4}&\color{blue}{12}&\color{orangered}{-33} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -33 }\right) $.