Synthetic division calculator

Example 1 : Divide $ x^2 +3x - 2 $ by $x - 2$.

Step 1: Write down the coefficients of $ 2x^2 +3x +4 $ into the division table.

$$ \begin{array}{c|rrr} \color{blue}{\square} &2&3&4\\ & & & \\ \hline & & & \end{array} $$

Step 2: Change the sign of a number in the divisor and write it on the left side. In this case, the divisor is $x - 2$ so we have to change $\, -2 \,$ to $\, \color{blue}{2} $.

$$ \begin{array}{c|rrr} \color{blue}{2} &2&3&4\\ & & & \\ \hline & & & \end{array} $$

Step 3: Carry down the leading coefficient

$$ \begin{array}{c|rrr} 2 &\color{orangered}{2}&3&4\\ & & & \\ \hline &\color{orangered}{2}& & \end{array} $$

Step 4: Multiply carry-down by left term and put the result into the next column

$$ \begin{array}{c|rrr} \color{blue}{2} &2&3&4\\ & &\color{blue}{4} & \\ \hline &\color{blue}{2}& & \end{array} $$

Step 5: Add the last column

$$ \begin{array}{c|rrr} 2 &2&\color{orangered}{3}&4\\ & &\color{orangered}{4}& \\ \hline &2&\color{orangered}{7}& \end{array} $$

Step 6: Multiply previous value by left term and put the result into the next column

$$ \begin{array}{c|rrr} \color{blue}{2} &2&3&4\\ & &4&\color{blue}{14} \\ \hline &2&\color{blue}{7}& \end{array} $$

Step 6: Add the last column

$$ \begin{array}{c|rrr} \color{blue}{2} &2&3&\color{orangered}{4}\\ & &4&\color{orangered}{14} \\ \hline &2&7& \color{orangered}{18} \end{array} $$

Step 7: Read the result from the synthetic table.

$$ \begin{array}{c|rrr} 2&2&3&4\\ & &4&14\\ \hline &\color{blue}{2}&\color{blue}{7}& \color{orangered}{18} \end{array} $$

The quotient is $ \color{blue}{2x + 7}$ and the remainder is $\color{orangered}{18}$.

Starting polynomial $ x^2 +3x - 2 $ can be written as:

$$ x^2 +3x - 2 = \color{blue}{2x + 7} + \dfrac{ \color{orangered}{18} }{ x - 2 } $$

Example 2 : Divide $ x^4 + 10x + 1 $ by $x + 2$.

Step 1: Write down the coefficients of $ x^4 - 10x + 1 $ into the division table. (Note that this polynomial doesn't have $x^3$ and $x^2$ terms, so these coefficients must be zero)

$$ \begin{array}{c|rrr} \color{blue}{\square} &1&0&0& 10&1\\ & & & & &\\ \hline & & & & & \end{array} $$

Step 2: Change the sign of a number in the divisor and write it on the left side. In this case, the divisor is $x + 3$ so we have to change $\, +3 \,$ to $\, \color{blue}{-3} $.

$$ \begin{array}{c|rrr} \color{blue}{-3}&1&0&0&10&1\\ & & & & &\\ \hline & & & & & \end{array} $$

Step 3: Carry down the leading coefficient

$$ \begin{array}{c|rrr} \color{blue}{-3}&\color{orangered}{1}&0&0&10&1\\ & & & & &\\ \hline &\color{orangered}{1}& & & & \end{array} $$

Multiply carry-down by left term and put the result into the next column

$$ \begin{array}{c|rrr} \color{blue}{-3}&1&0&0&10&1\\ & &\color{blue}{-3}& & &\\ \hline &\color{blue}{1}& & & & \end{array} $$

ADD the last column

$$ \begin{array}{c|rrr} -3 &1&\color{orangered}{0}&0&10&1\\ & &\color{orangered}{-3}& & &\\ \hline &1&-3 & & & \end{array} $$

Multiply last value by left term and put the result into the next column

$$ \begin{array}{c|rrr} \color{blue}{-3} &1&0&0&10&1\\ & &-3&\color{blue}{9}& &\\ \hline &1&\color{blue}{-3} & & & \end{array} $$

ADD the last column

$$ \begin{array}{c|rrr} -3 &1& 0&\color{orangered}{0}&10&1\\ & &-3&\color{orangered}{9}& &\\ \hline &1&-3&\color{orangered}{9}& & \end{array} $$

Multiply last value by left term and put the result into the next column

$$ \begin{array}{c|rrr} \color{blue}{-3} &1& 0&0&10&1\\ & &-3&9& \color{blue}{-27}&\\ \hline &1&-3&\color{blue}{9}& & \end{array} $$

ADD the last column

$$ \begin{array}{c|rrr} -3 &1&0&0&10&\color{orangered}{1}\\ & &-3& 9 & \color{orangered}{-27}&\\ \hline &1&-3&9& \color{orangered}{-17}& \end{array} $$

Multiply last value by left term and put the result into the next column

$$ \begin{array}{c|rrr} \color{blue}{-3} &1&0&0&10&1\\ & &-3& 9 &-27&\color{blue}{51}\\ \hline &1&-3&9&\color{blue}{-17}& \end{array} $$

ADD the last column

$$ \begin{array}{c|rrr} -3 &1&0&0&10&\color{orangered}{1}\\ & &-3& 9 &-27&\color{orangered}{51}\\ \hline &1&-3&9&-17&\color{orangered}{52} \end{array} $$

Step 7: Read the result from the synthetic table.

$$ \begin{array}{c|rrr} -3 &1&0&0&10&\color{orangered}{1}\\ & &-3& 9 &-27&\color{orangered}{51}\\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{9}&\color{blue}{-17}&\color{orangered}{52} \end{array} $$

The quotient is $ \color{blue}{x^3 - 3x^2 + 9x - 17}$ and the remainder is $\color{orangered}{52}$.

Starting polynomial $ x^4 + 10x + 1 $ can be written as:

$$ x^4 + 10x + 1 = \color{blue}{x^3 - 3x^2 + 9x - 17} + \dfrac{ \color{orangered}{52} }{ x + 3 } $$