How to factorize? ;factorization methods/types and examples

First of all, what is factorization?

It is the method of writing numbers as the product of their factors or divisors.
Now let's go to the methods of factorizing various types of polynomials in detail.

Type 1. Monomial Factors
The basis for multiplication of polynomials is the distributive property of numbers. This property states:

This shows how, a sum of two terms, which have a factor in common, can be expressed as a product. Thus, the factor a in each term on the left side can be taken out as a common factor of the whole expression and we can write ab + ac = a(b + c); here a and b+c are the factors of ab+ac. A factor such as a is a common monomial factor in the terms ab and ac.
Examples;
Consider 21 = 3 * 7; here 3 and 7 are the factors of 21.
Let's take a complex one; 28x³ - 70x².
Now this can be factorized in the following ways:  
14(2x³ - 5x²), x(28x² - 70x), x²(28x - 70),etc.

Type 2. When the Common Factor is a Polynomial
Form: x(a + b) + y(a + b) + z(a + b)
In ab + ac = a(b + c). for all a,b,c.
You may replace a,b,c by expression involving more than 1 term;
For example;
(x + 5)x+(x + 5)y = (x + 5)(x + y)

Type 3. By grouping the Terms
Form: ax + ay + bx + by
Some polynomials with four terms that do not have a common factor in each term, but do have some similar terms, such as 3m + 3n - an - am, can be factorized by grouping. This means grouping in a manner that will express the given polynomial in the form of a binomial with a common binomial factor.
The first step in this process is to separate the given terms into binomial groups having a common factor to each group. Thus, 

3m + 3n - an - am = (3m + 3n) + (- am - an)

When this can be done, each binomial in the expression may be factorized as a binomial with a common monomial factor. That is,

(3m + 3n) + (-am - an) = [(3(m + n)] + [-a(m + n)] = (m + n)(3 - a)

Notice that a polynomial factorable by grouping has 

1. 4 terms

2. can be grouped into two binomials with a common factor in each

3. such that the binomials factor in each group is the same.

Example;
Let's factorize ax + bx + ay + by
sol. ax + bx + ay + by
= (ax + bx) + (ay + by)
= x(a + b) + y(a + b)
= (a + b)(x + y)

Type 4. Factorising Perfect Square Trinomials

Form: a² + 2ab + b²   and a² -2ab + b²

(Trinomials is which two terms are perfect squares and the third term is twice the product of the square roots of these two perfect square terms).

Formulas:
a² + 2ab + b² = (a + b)²
a² - 2ab + b²  = (a - b)²

Example;
Factorise 36x² +60xy + 25y²

= (6x)² + 2(6x)(5y) + (5y)²

= (6x + 5y)²

= (6x + 5y)(6x + 5y)

Type 5. Difference of two squares
Form: a² - b² = (a + b)(a - b)

You must be familiar with products of the following type:

(a+b)(a-b) = a² - b² 
(3x + y)(3x - y) = 9x² - y²
(4x - 7y)(4x + 7y) = 16x² - 49y²

Since in each expression on the right hand side, two perfect squares are separated by a subtraction sign, therefore, each of these expression is referred to as the differences of two squares.

You can reverse this pattern to find a pattern for factorising the difference of two squares.

Since the first term of the difference of two squares is the square of some expression, the first term of each factor must be the square root of the first square.

Similarly, the second term of each factor is the square root of the second square. One factor must have ‘ + ‘ between the terms, and the other must have ‘ – ‘.

Example;

Factorise 4x² - 9y²

Sol. Now this can be split into 3 steps:

1.      This is the difference of two squares, so you find the square root of the first term. (2x)(2x)

2.      Next, you find the square root of the second term. (3y)(3y)

3.      Now fill in the signs. One has a plus sign, the other has a minus sign. Although it does not matter which comes first, you must have one of each. (2x – 3y)(2x + 3y)