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

The synthetic division table is:

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

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

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.

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

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

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

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

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

Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 0 } = \color{orangered}{ 5 } $

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

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.

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

Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 0 } = \color{orangered}{ -4 } $

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

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 0 } $.

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

Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 0 } = \color{orangered}{ -5 } $

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

Remainder is the last entry in the bottom row $ \left(\color{red}{ -5 }\right) $.

Synthetic Division Calculator