Find remainder when $ 4x^{3}-12x^{2}-5x-1 $ is divided by $ 2x+1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&4&-12&-5&-1\\& & -4& 16& \color{black}{-11} \\ \hline &\color{blue}{4}&\color{blue}{-16}&\color{blue}{11}&\color{orangered}{-12} \end{array} $$The remainder when $ 4x^{3}-12x^{2}-5x-1 $ is divided by $ x+1 $ is $ \, \color{red}{ -12 } $.
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 + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-12&-5&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 4 }&-12&-5&-1\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 4 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-12&-5&-1\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}-1&4&\color{orangered}{ -12 }&-5&-1\\& & \color{orangered}{-4} & & \\ \hline &4&\color{orangered}{-16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-12&-5&-1\\& & -4& \color{blue}{16} & \\ \hline &4&\color{blue}{-16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 16 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}-1&4&-12&\color{orangered}{ -5 }&-1\\& & -4& \color{orangered}{16} & \\ \hline &4&-16&\color{orangered}{11}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 11 } = \color{blue}{ -11 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-12&-5&-1\\& & -4& 16& \color{blue}{-11} \\ \hline &4&-16&\color{blue}{11}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -11 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}-1&4&-12&-5&\color{orangered}{ -1 }\\& & -4& 16& \color{orangered}{-11} \\ \hline &\color{blue}{4}&\color{blue}{-16}&\color{blue}{11}&\color{orangered}{-12} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -12 }\right) $.