Find remainder when $ 5x^{3}-6x^{2}-7x+3 $ is divided by $ x+4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&5&-6&-7&3\\& & -20& 104& \color{black}{-388} \\ \hline &\color{blue}{5}&\color{blue}{-26}&\color{blue}{97}&\color{orangered}{-385} \end{array} $$The remainder when $ 5x^{3}-6x^{2}-7x+3 $ is divided by $ x+4 $ is $ \, \color{red}{ -385 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&-6&-7&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 5 }&-6&-7&3\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 5 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&-6&-7&3\\& & \color{blue}{-20} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -26 } $
$$ \begin{array}{c|rrrr}-4&5&\color{orangered}{ -6 }&-7&3\\& & \color{orangered}{-20} & & \\ \hline &5&\color{orangered}{-26}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -26 \right) } = \color{blue}{ 104 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&-6&-7&3\\& & -20& \color{blue}{104} & \\ \hline &5&\color{blue}{-26}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 104 } = \color{orangered}{ 97 } $
$$ \begin{array}{c|rrrr}-4&5&-6&\color{orangered}{ -7 }&3\\& & -20& \color{orangered}{104} & \\ \hline &5&-26&\color{orangered}{97}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 97 } = \color{blue}{ -388 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&-6&-7&3\\& & -20& 104& \color{blue}{-388} \\ \hline &5&-26&\color{blue}{97}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -388 \right) } = \color{orangered}{ -385 } $
$$ \begin{array}{c|rrrr}-4&5&-6&-7&\color{orangered}{ 3 }\\& & -20& 104& \color{orangered}{-388} \\ \hline &\color{blue}{5}&\color{blue}{-26}&\color{blue}{97}&\color{orangered}{-385} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -385 }\right) $.