Factorisation

What is Factorisation?

When we break a number or a polynomial into a product of many factors of other polynomials, which when multiplied gives the original number, it is called factorisation. 

For example -  Prime Factorisation of 28 = 2 X 2 X 7 and

Product of 2 * 7 * 7 = 28 (Original Number).

There are only two types of numbers: Prime number and Composite number. 

Before starting factorisation, let us first discuss an important mathematical term ‘Factor’.

What is Factor? 

Factors are the numbers, algebraic variables or an algebraic expression which divides the number or an algebraic expression without leaving any remainder.

For Example: 

  1. The factors of 42 = 2 × 3 × 7, the numbers 2, 3 and 7 are the factors of 42 and divide it without leaving any remainder.

  2. The factors of an algebraic expression 10xyz = 2 × 5 × x × y × z.

Similarly,

  1. The algebraic expression 5(x + 1) (z + 2) can be written in irreducible factor form as:

5(x + 1) (z + 2) = 5 × (x + 1) × (z + 2)

What is Factorisation?

The factorisation is defined as expressing or decomposing a number or an algebraic expression as a product of its prime factors or irreducible factors.

In this article, we will learn how to factorise an algebraic expression as products of its factors.

From previous examples of algebraic expressions 10xyz and 5(x + 1) (z + 2), their factors can be easily obtained by just reading off them.

Now consider the algebraic expressions, 3y+12xy, 5x+10, x2 + x y + 8x and x2+2x+1, it is not so easy to tell their factors. To find factors of these expressions, we need to follow some methods of factorisation

Methods of Factorisation

  1. Method of common factors


Step 1: Each term of given algebraic expression is written as a product of irreducible factors.

Step 2: The common factors are taken out and the rest of the expression is combined in the brackets.


For example 1: Factorise 6xy + 15yz


Step 1:  we have, 6xy = 2 × 3 × x × y

                            15yz = 3 × 5 × y × z

Step 2: Common factors of these terms are 3 and y.

Therefore, 6xy + 15yz = (2 × 3 × x × y) + (3 × 5 × y × z)

                  6xy + 15yz = 3y (2x + 5z)


  1. Factorisation by Regrouping


Sometimes, all the terms of a given expression do not have a common factor. 

But these terms can be grouped in such a way that all the terms in each group have a common factor.

When we do this, a common factor comes out from all the groups and leads to the required factorisation of the expression. 


For example 2: Factorise 3x2y - 9x + 4xy – 12


Since, there is no common factor among the terms of given expression. So, we will group them separately to get a common factor from these groups.


We observe that the terms 3x2y - 9x have a common factor of 3x and the terms 4xy – 12 have a common factor of 4.


i.e.                  3x2y - 9x = 3x (xy - 3) and 4xy – 12 = 4 (xy – 3)


Putting these grouped terms together, we get:


3x2y - 9x + 4xy – 12 = 3x (xy – 3) + 4 (xy – 3)

                            = (3x + 4) (xy – 3)

Therefore, the factors of 3x2y - 9x + 4xy – 12 are (3x + 4) and (xy – 3).


  1. Factorisation using identities


A number of algebraic expressions can be factorised by putting them in the form of suitable identities. These identities are: 


  1. a2 + 2ab + b2 = (a + b)2 


  1. a2 – 2ab + b2 = (a – b)2 


  1. a2 – b2 = (a + b) (a – b)


For example3: Factorise x2 + 10x + 25


Above algebraic expression x2 + 10x + 25 matches with the form of identity a2 + 2ab + b2

Where, a = x and b = y

Such that, a2 + 2ab + b2 = x2 + 2(x)(y) + 52


                x2 + 2(x)(y) + 52 = (x + 5)2  

Therefore, the factors of x2 + 10x + 25 are (x + 5) and (x + 5).


For example 4: Factorise 9x2 – 25y2

Above algebraic expression 9x2 – 25y2 matches with the form of identity a2 – b2

Where, a = 3x and b = 5y

Such that, a2 – b2 = (3x)2 – (5y)2

                 

                  9x2 – 25y2 = (3x)2 – (5y)2

                                   = (3x + 5y) (3x – 5y)

Therefore, the factors of 9x2 – 25y2 are (3x + 5y) and (3x – 5y).


  1. Factorising algebraic expression of form x2 + px + q


For factorising an algebraic expression of the form x2 + px + q, we find two factors a and b of the constant term i.e. q such that:

Product of a and b i.e. ab = q 

And sum of a and b i.e.  a + b = p


Then, the given algebraic expression becomes:

    x2 + (a + b) x + ab 

x2 + ax + bx + ab 

x (x + a) + b (x + a) 

(x + a) (x + b)  which are the required factors of given algebraic expression.


For example 5: Factorise x2 + 8x + 15

            First of all we will factorise the constant term 15 into two factors such that 

            their Sum = 8 and 

            Product = 15

            Clearly, the numbers are 5 and 3

            So we can write, x2 + 8x + 15 = x2 + 5x + 3x + 15

                                                   = x (x + 5) + 3(x + 5)

                                                   = (x + 5) (x + 3)

Therefore, the factors of x2 + 8x + 15 are (x + 5) and (x + 3).