Polynomial factoring calculator

This calculator is a free online math tool that writes a polynomial in factored form.
The solver shows a complete step-by-step explanation.

problem

Factor polynomial $ 2a-4b+a^{2}-2ab $.

solution

$$ 2a-4b+a^{2}-2ab = (2+a)(a-2b) $$

explanation

To factor $ 2a-4b+a^2-2ab $ we can use factoring by grouping.

Group $ \color{blue}{ 2a }$ with $ \color{blue}{ -4b }$ and $ \color{red}{ a^2 }$ with $ \color{red}{ -2ab }$ then factor each group.

$$ \begin{aligned} 2a-4b+a^2-2ab &= ( \color{blue}{ 2a-4b } ) + ( \color{red}{ a^2-2ab }) = \\ &= \color{blue}{ 2( a-2b )} + \color{red}{ a( a-2b ) } = \\ &= (2+a)(a-2b) \end{aligned} $$

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Polynomial Factoring Calculator (shows all steps)
supports polynomials with both single and multiple variables
show help ↓↓ examples ↓↓ tutorial ↓↓
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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 $a^2 + 2a + 1 - b^2$
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TUTORIAL

Polynomial Factoring Techniques

To find the factored form of a polynomial, this calculator employs the following methods:

1. Factoring GCF, 2 Factoring by grouping, 3 Using the difference of squares, and 4 Factoring Quadratic Polynomials

Method 1 : Factoring GCF

Example 01: Factor $ 3ab^3 - 6a^2b $

$$ \begin{aligned} 3ab^3 - 6a^2b &= \color{blue}{3 \cdot a \cdot b} \cdot b \cdot b - 2 \cdot \color{blue}{3 \cdot a } \cdot a \color{blue}{\cdot b} = \\[1 em] &= \color{blue}3ab(b^2-2a) \end{aligned} $$

Method 2 : Factoring By Grouping

The method is very useful for finding the factored form of the four term polynomials.

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

We usually group the first two and the last two terms.

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

We now factor $ \color{blue}{2} $ out of the blue terms and $ \color{red}{a} $ out of 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 the first two terms of a polynomial do not have a common factor, so we have to group the first and third terms 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 method 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 the 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, thus the calculation will be similar to the last 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 the 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 by using an example.

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

We know that the factored form has the following pattern

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

All we have to do now is fill in the blanks with the two numbers. To do this, notice that the product of these two numbers has to be 4 and their sum has to be 5. We conclude, after some trial and error, that the missing numbers are $ \color{blue}{1} $ and $ \color{red}{4} $. As a result, the solution is::

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

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

Like in the 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.

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