Find remainder when $ 3x^{4}-16x^{3}-7x^{2}+64x-20 $ is divided by $ x+4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&3&-16&-7&64&-20\\& & -12& 112& -420& \color{black}{1424} \\ \hline &\color{blue}{3}&\color{blue}{-28}&\color{blue}{105}&\color{blue}{-356}&\color{orangered}{1404} \end{array} $$The remainder when $ 3x^{4}-16x^{3}-7x^{2}+64x-20 $ is divided by $ x+4 $ is $ \, \color{red}{ 1404 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&-16&-7&64&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 3 }&-16&-7&64&-20\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&-16&-7&64&-20\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -28 } $
$$ \begin{array}{c|rrrrr}-4&3&\color{orangered}{ -16 }&-7&64&-20\\& & \color{orangered}{-12} & & & \\ \hline &3&\color{orangered}{-28}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -28 \right) } = \color{blue}{ 112 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&-16&-7&64&-20\\& & -12& \color{blue}{112} & & \\ \hline &3&\color{blue}{-28}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 112 } = \color{orangered}{ 105 } $
$$ \begin{array}{c|rrrrr}-4&3&-16&\color{orangered}{ -7 }&64&-20\\& & -12& \color{orangered}{112} & & \\ \hline &3&-28&\color{orangered}{105}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 105 } = \color{blue}{ -420 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&-16&-7&64&-20\\& & -12& 112& \color{blue}{-420} & \\ \hline &3&-28&\color{blue}{105}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 64 } + \color{orangered}{ \left( -420 \right) } = \color{orangered}{ -356 } $
$$ \begin{array}{c|rrrrr}-4&3&-16&-7&\color{orangered}{ 64 }&-20\\& & -12& 112& \color{orangered}{-420} & \\ \hline &3&-28&105&\color{orangered}{-356}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -356 \right) } = \color{blue}{ 1424 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&-16&-7&64&-20\\& & -12& 112& -420& \color{blue}{1424} \\ \hline &3&-28&105&\color{blue}{-356}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 1424 } = \color{orangered}{ 1404 } $
$$ \begin{array}{c|rrrrr}-4&3&-16&-7&64&\color{orangered}{ -20 }\\& & -12& 112& -420& \color{orangered}{1424} \\ \hline &\color{blue}{3}&\color{blue}{-28}&\color{blue}{105}&\color{blue}{-356}&\color{orangered}{1404} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 1404 }\right) $.