Polynomial factoring calculator

Method 1 : Factoring GCF

Example 01: Factor 3ab³-6a²b

3ab3-6a2b=3·a·b·b·b-2·3·a·a·b=
         =3ab(b2-2a)

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+a²-2ab

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

2a-4b+a2-2ab = 2a-4b + a2-2ab

We now factor 2 out of the blue terms and a out of from red ones.

2a-4b+a2-2ab = 2·(a-2b)+a·(a-2b)

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

2·(a-2b)+a·(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.

5ab+2b+5ac+2c = 5ab+5ac+2b+2c=
              = 5a(b+c) + 2(b+c)
              = (b+c)(5a + 2)

Method 3 : Special Form – A

The most common special case is the difference of two squares.

a²-b²=(a+b)(a-b)

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

Example 05: Factor 4x²-y².

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

4x²-y²=(2x)²-y²

Now we can apply above formula with a=2x and b=y

(2x)²-y²=(2x-y)(2x+y)

Example 06: Factor 9a²b⁴-4c²

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

9a2b4-4c2=(3ab2)2-4c2=
         =(3ab2-2c)(3ab2+2c)

Method 4 : Special Form – B

The second special case of factoring is the Perfect Square Trinomial

a² ± 2ab + b² = (a ± b)²

Example 07: Factor 100+20x+x²

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

100+20x+x2 = 102+2·10·x+x2 =
           = (10+x)2

Method 5: Factoring Quadratic Polynomials

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

Example 08: Factor x² + 3x + 4

We know that the factored form has the following pattern

x2+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 1 and 4. As a result, the solution is::

x^2+5x+4 = (x+1)(x+4)

Example 09: Factor x² - 8x + 15

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

x2-8x+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 is not hard to see that two numbers with such properties are -3 and -5, so the solution is.

x2+5x+4 = (x+-3)(x+-5) =
    = (x-3)(x-5)