Find remainder when $ 4x^{3}-12x^{2}+5x-2 $ is divided by $ 2 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}0&4&-12&5&-2\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{5}&\color{orangered}{-2} \end{array} $$

The remainder when $ 4x^{3}-12x^{2}+5x-2 $ is divided by $ x $ is $ \, \color{red}{ -2 } $.

We can find remainder using synthetic division method.

Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.

$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-12&5&-2\\& & & & \\ \hline &&&& \end{array} $$

Step 1 : Bring down the leading coefficient to the bottom row.

$$ \begin{array}{c|rrrr}0&\color{orangered}{ 4 }&-12&5&-2\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-12&5&-2\\& & \color{blue}{0} & & \\ \hline &\color{blue}{4}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $

$$ \begin{array}{c|rrrr}0&4&\color{orangered}{ -12 }&5&-2\\& & \color{orangered}{0} & & \\ \hline &4&\color{orangered}{-12}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-12&5&-2\\& & 0& \color{blue}{0} & \\ \hline &4&\color{blue}{-12}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 0 } = \color{orangered}{ 5 } $

$$ \begin{array}{c|rrrr}0&4&-12&\color{orangered}{ 5 }&-2\\& & 0& \color{orangered}{0} & \\ \hline &4&-12&\color{orangered}{5}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-12&5&-2\\& & 0& 0& \color{blue}{0} \\ \hline &4&-12&\color{blue}{5}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 0 } = \color{orangered}{ -2 } $

$$ \begin{array}{c|rrrr}0&4&-12&5&\color{orangered}{ -2 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{5}&\color{orangered}{-2} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ -2 }\right) $.

Synthetic Division Calculator