Polynomial factoring calculator

Tutorial

What is polynomial factoring?

Polynomial factorization expresses a polynomial as a product of two or more polynomials whose degrees are smaller than the original polynomial's.

Why is polynomial factoring important?

Factoring polynomials is useful when we need to simplify algebraic fractions or when we want to solve polynomial equations. Thus, it is much easier to solve various types of math problems related to polynomials, such as equations, inequalities and system of equations, once we know how to factor polynomials.

How we can factor a polynomial

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 4. Perfect Square Trinomial 5. Factoring quadratic polynomials.

Examples

Method 1 : Difference of squares

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.

Learn how to prove difference of squares formula in 40 seconds.

Example 01: 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 02: 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 2 : Factoring GCF

Example 03: Factor 3ab³-6a²b

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

Method 3 : 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 4 : Perfect Square Trinomial

The second special case of factoring is the Perfect Square Trinomial.

a² ± 2ab + b² = (a ± b)²

Example 05: 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 06: 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 07: 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)