Points to Remember- Factorisation | Mathematics (Maths) Class 8 PDF Download

What are Factors?

Factors are components that, when multiplied, form a given expression or number. Factorization breaks down complex expressions into simpler parts for analysis or manipulation.

Algebraic Identities to Remember

• Factorisation means to write an expression as a product of its factors.
• Prime factor, an irreducible factor, a factor which cannot be expressed further as a product of factors.
• Some expressions can easily be factorized using these identities:
I. a² + 2ab + b² = (a + b)²
II. a² – 2ab + b² = (a – b)²
III. a² – b² = (a – b)(a + b)
IV. x² + (a + b)x + ab = (x + a)( x+ b)
• The number 1 is a factor of every algebraic term, but it is shown only when needed.
• When factorisation of x² + (a + b)x + ab is done by splitting the middle term, the two numbers which give the product ab and (a + b) as the coefficient of x have to be chosen very carefully with correct sign.

Note: 
In case of factorisation of a term of an expression, the word ‘irreducible’ is used in place of ‘prime’. For example, 6pq = 2 * 3 * pq is not the irreducible form because pq can further be factorised as p q, i.e. the irreducible form of 6pq = 2 * 3 * p * q.

Example: Write 10y as irreducible factor form.

Solution: We have  

10 = 2 * 5

xy = x *y

∴  10xy = 2 * 5* x * y

Methods of Factorization

1. Factorization Using Common Factors

To factorize an algebraic expression, the highest common factors are determined.

Example 1: Algebraic expression -2y² + 8y can be written as 2y(-y+4), where 2y is the highest common factor in the expression.

Example 2: Factorise 18x² – 14x³ + 10x⁴

Solution: We have  

Obviously, the common factors of these terms are 2, and two times x.

∴ 18x² – 14x³ + 10x⁴

Thus, 18x² – 14x³ + 10x⁴ = 2x²[9 – 7x + 5x²]

2. Factorisation By Regrouping Terms

In some algebraic expressions, it is not possible that every term has a common factor. Therefore, to factorise those algebraic expressions, terms having common factors are grouped together. 

Example 1: Factorise 9x + 18y + 6xy + 27

Solution: Here, we have a common factor 3 in all the terms.

∴ 9x + 18y + 6xy + 27 = 3[3x + 6y + 2xy + 9]

We find that 3x + 6y = 3(x + 2y) and 2xy + 9 = 1(2xy + 9)

i.e. a common factor in both groups does not eist,

Thus, 3x + 6y + 2xy + 9 cannot be factorized.

On regrouping the terms, we have

3x + 6y + 2xy + 9 = 3x + 9 + 2xy + 6y

= 3(x + 3) + 2y(x + 3)

= (x + 3)(3 + 2y)

Now, 3[3x + 6y + 2xy + 9] = 3[(x + 3)(3 + 2y)]

Thus, 9x + 18y + 6xy + 27 = 3(x + 3)(2y + 3)

Example 2: Factorise 12a + n – na – 12

Solution:

= 12a-12+n-na

= 12(a-1)-n(a-1)

= (12-n)(a-1)

(12-n) and (a-1)are factors of the expression 12a+n-na-12

Some Solved Examples for You:

Q1: Let f(x)=2x³+16x²+44x+42 be a polynomial having one of the factors as (x²+5x+7), then the other factor of f(x) would be a multiple of:
A) 1
B) 2
C) 3
D) 4

Solution: B) Since f(x) is a cubic polynomial, and one of the factors is a polynomial of degree two, then we can say that the other factor will be a polynomial of the form ax + b; where ‘a’ nd ‘b’ are two constants and a ≠ 0. Hence, we can write:

2x³+16x²+44x+42
 = (x²+5x+7) × (ax + b)
= ax³ + bx² + 5ax² + 5bx + 7ax + 7b

or 2x³+16x²+44x+42
 = ax³ + (b + 5a)x² +x² + (5b + 7a)x + 7b
Comparing the coefficients of x on both sides, we have 2 = a and 42 = 7b. Therefore, b = 6 and a = 2. hence the other factor is 2x + 6 or 2(x+3) which is a multiple of 2.

Q 2: Factorise: 5m2 − 8m − 4:
A) (5m + 2)(m + 2)
B) (5m – 2)(m – 2)
C) (5m – 2)(m + 2)
D) (5m + 2)(m – 2)

Solution: D) The given expression is: 5m2 – 8m – 4. Therefore, it can be written as:

5m2 – 10m + 2m – 4 = 5m(m – 2) + 2(m –  2)

Hence we can write this = (5m + 2)(m – 2)