- All Math Calculators
- ::
- Polynomial Calculators
- ::
- Polynomial Factoring Calculator

This calculator is free online math tool that writes a polynomial in a factored form.

The solver shows a complete step-by-step explanation.

working...

EXAMPLES

example 1:ex 1:

factor $x^2 - 10x + 25$

example 2:ex 2:

factor $3x^2 - 27$

example 3:ex 3:

factor $4x^3 - 21x^2 + 29x -6$

example 4:ex 4:

factor $3uv-12u+6v-24 $

example 5:ex 5:

factor $a^2 + 2a + 1 - b^2$

TUTORIAL

This calculator applies nine different methods to factor polynomials, but the most common ones are five of them, which are described below.

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

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

$$ \begin{aligned} 5ab + 2b + 5ac + 2c &= \color{blue}{5ab + 5ac} + \color{red}{2b + 2c} = \\ &=\color{blue}{5a (b + c)} + \color{red}{2 (b + c)} = \\ &= (b+c)(\color{blue}{5a} + \color{red}{2}) \end{aligned} $$The most common special case is the **difference of two squares**

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.

$$ 9a^2b^4 - 4c^2 = \left( 3ab^2 \right)^2 - \left( 2c \right)^2 = (3ab^2-2c)(3ab^2+2c) $$The second special case of factoring is the **Perfect Square Trinomial**

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

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

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

**Quick Calculator Search**

** Please tell me how can I make this better.**

222 529 319 solved problems