Polynomial factoring calculator

Method 1 : Greatest Common Factor (GCF)

This is always the first method we should try when factoring polynomials.

Example 01: Factor $ 10a^2b + 15ab^2 $ using GCF method.

Here we can factor out $ \color{blue}{ 5ab } $ of both terms.

$$ \begin{aligned} 10a^2b + 15ab^2 & = \color{blue}{5} \cdot \color{red}{2} \cdot \color{blue}{ab} \cdot \color{red}{a} + \color{blue}{5} \cdot \color{red}{3} \cdot \color{blue}{ab} \cdot \color{red}{b} = \\ &= \color{blue}{5ab} \, ( \color{red}{2a} + \color{red}{3b}) \end{aligned} $$

Example 02: Factor $ 20x^2y^3 + 30x^2y + 40x^2y^2$.

Notice that each term contains $\color{blue}{10 x^2 y}$, as common factor

$$ \begin{aligned} 20x^2y^3 &+ 30x^2y + 40x^2y^2 = \\ &= \color{blue}{10x^2y} \cdot \color{red}{2y^2} + \color{blue}{10x^2y} \cdot \color{red}{1} + \color{blue}{10x^2y} \cdot \color{red}{y} = \\ &= \color{blue}{10x^2y} \, \left( \color{red}{2y^2} + \color{red}{1} + \color{red}{y} \right) \end{aligned} $$

Method 2 : Factoring By Grouping

The method is very useful to find the factored form of the four term polynomials.

Example 03: Factor $ 2a - 4b + a^2 - 2ab $

We usually group first two and the last two terms.

$$ 2a - 4b + a^2 - 2ab = \color{blue}{2a - 4b} + \color{red}{a^2 - 2ab} $$

Now we factor $ \color{blue}{2} $ from the blue terms and $ \color{red}{a} $ from red ones.

$$ \color{blue}{2a - 4b} + \color{red}{a^2 - 2ab} = \color{blue}{2 \cdot (a - 2b)} + \color{red}{a \cdot (a - 2b) } $$

At the end we factor out common factor of $ (a - 2b) $

$$ 2 \cdot \color{blue}{(a - 2b)} + a \cdot \color{blue}{(a - 2b)} = (a - 2b) (2 + a) $$

Example 04: Factor $ 5ab + 2b + 5ac + 2c $

This is a rare situation where first two terms of polynomial do not have a common factor. so we have to group the first and third term together.

$$ \begin{aligned} 5ab + 2b + 5ac + 2c &= \color{blue}{5ab + 5ac} + \color{red}{2b + 2c} = \\ &=\color{blue}{5a (b + c)} + \color{red}{2 (b + c)} = \\ &= (b+c)(\color{blue}{5a} + \color{red}{2}) \end{aligned} $$

Method 3 : Special Form - A

The most common special case is the difference of two squares

$$ a^2 - b^2 = (a+b) (a-b) $$

We usually use this formula when the polynomial has only two terms.

Example 05: Factor $ 4x^2 - y^2 $.

First we need to notice that the polynomial can be written as difference of two perfect squares.

$$ \color{blue}{4x^2} - \color{red}{y^2} = \color{blue}{ \left(2x\right)}^2 - \color{red}{y}^2 $$

Now we can apply above formula with $ \color{blue}{a = 2x} $ and $ \color{red}{b = y} $

$$ \left( \color{blue}{2x} \right)^2 - \color{red}{y}^2 = (2x - b) (2x + b) $$

Example 06: Factor $ 9a^2b^4 - 4c^2 $.

The binomial we have here is the difference of two perfect squares, so the the calculation will be similar to previous one.

$$ 9a^2b^4 - 4c^2 = \left( 3ab^2 \right)^2 - \left( 2c \right)^2 = (3ab^2-2c)(3ab^2+2c) $$

Method 4 : Special Form - B

The second special case of factoring is the Perfect Square Trinomial

$$ a^2 \pm 2ab + b^2 = (a \pm b)^2 $$

Example 07: Factor $ 100 + 20x + x^2 $

In this case the first and third terms are perfect squares. So we can use above formula.

$$ \color{blue}{100} + 20x + \color{red}{x^2} = \color{blue}{10}^2 + 2 \cdot \color{blue}{10} \cdot \color{red}{x} + \color{red}{x}^2 = (10 + x)^2 $$

Method 5: Factoring Quadratic Polynomials

The best way to explain this method is using an example.

Example 08: Factor $ x^2 + 3x + 4 $

We know that the factoring has the following form

$$ x^2 + 5x + 4 = (x + \_ ) (x + \_ ) $$

We only have to find the two numbers in empty spots. To do this, notice that the product of these two numbers has to be 4 and their sum have to be 5. After some trial and error, we conclude that the missing numbers are $ \color{blue}{1} $ and $ \color{red}{4} $. So the solution is:

$$ x^2 + 5x + 4 = (x + \color{blue}{1} ) (x + \color{red}{4} ) $$

Example 09: Factor $ x^2 - 8x + 15 $

Like the in previous example, we look again for the solution in the form

$$ x^2 - 8x + 15 = (x + \_ ) (x + \_ ) $$

The sum of missing numbers is $-8$ so we need to find two negative numbers such that the product is $15$ and the sum is $-8$. It's not hard to see that two numbers with such properties are $-3$ and $-5$, so the solution is.

$$ x^2 + 5x + 4 = (x + \color{blue}{-3} ) (x + \color{red}{-5} ) = (x-3)(x-5) $$