Notes of chapter: Factorisation are presented below. Indepth notes along with worksheets and NCERT Solutions for Class 8.
(i) The prime factors of a natural number are prime numbers when multiply together we will get that number.
Eg.- Find the prime factors of 40.
Ans-
40 = 2 × 2 × 2 × 5
Hence, 2 and 5 are prime factors of 40.
(ii) The factors of an algebraic expression are terms when multiply together we will get that algebraic expression.
Eg:-Find the factors of 2x y.
Ans-
It is an algebraic expression.
2xy is a term.
Factors of term 2xy =2,x,y
[Note – 1 is factor of every term. We do not write 1 as a separate factor unless it is specially required.]
Types of factors of algebraic expressions-
(a) Irreducible factors of algebraic expressions are the factors which can not be factorised further.
Eg:-
(A) Find the irreducible factors of 2xy.
Irreducible factors of 2xy =2,x,y
2, x and y all are irreducible factors because we can not reduce them.
(B) Find the irreducible factors of 5x(y + 2).
Irreducible factors of 5x(y + 2) = 5 × x × (y + 2)
5,x and (y + 2 ) are irreducible factors.
(b) Reducible factors of algebraic expressions are the factors which can be factorised further.
Eg:- 2xy is a reducible factors.
(2)Factorisation is a method to do the factors as a product of the number or algebraic expression.
Eg:-
(i) Factorise a number 35.
Ans-
35 = 5 × 7
Hence, 1, 5, 7and 35 are factors of 35.
(ii)Factorise an algebraic expression 4 x2.
4 x2 = 2 × 2 × x × x
Hence, 1, 2, x are factors of 4x2.
Methods of factorisation
(i) Method of common factors
(a)Factorise 3x + 9
Step 1-
Find irreducible factors of each term.
3x = 3 × x
9 = 3 × 3
Therefore,
3x + 9 = (3× x) + (3 × 3)
Step 2-
Find the common factor of terms.
In above factors only 3 is a common factor.
Step 3-
By using distributive law, take out the common factors of factors.
(3× x) + (3 × 3) = 3(x + 3)
Step 4-
Hence,
3x + 9 = 3(x + 3)
3 and (x + 3) are the factors of the expression 3x + 9.
(b) Factorise 3xy + 9x
Step 1 –
Find irreducible factors of each term.
3xy = 3 × x × y
9x = 3× 3 × x
Therefore,
3xy + 9x = (3 × x × y) + (3 × 3 × x)
Step 2-
Find the common factor of terms.
In above factors only 3 and x are common factors.
Step 3-
By using distributive law, take out the common factors of terms.
(3× x × y) + (3× 3× x) = 3x (y + 3)
Step 4-
Hence,
3xy + 9x = 3x(y + 3)
3x and (y + 3) are the factors of the expression 3x y+ 9x.
(ii) Method of regrouping terms
Regrouping means rearranging the terms in different groups in such a way that each group must has one common factor.
Eg:- Find the factor of expression 4x y + 5 + 2x + 10y
Step 1-
Regroup the expressions in such a way that each group must have one common factor.
4x y + 5 + 2x + 10y
Regroup- (4xy + 2x )+ (5 + 10y)
Step 2 –
Find irreducible factors of each term.
4xy = 2 × 2 × x × y
2x = 2×x
10y = 5 × 2 × y
5 = 5
Therefore
4xy + 2x + 5 + 10y = (4 x y + 2 x) + (5 + 5 × 2 × y)
(iii) Factorisation using identities
(a) (a + b)2 = a 2+ 2ab + b2
(b) (a – b)2 = a 2– 2ab + b2
(c) (a + b) (a – b) = a2– b2
(d) (x + a ) (x + b) = x2 + (a + b)x + ab
Steps to use identities-
(a) (a + b)2 = a 2+ 2ab + b2
Eg:- Factorise x2 + 6x + 9
Ans-
x2 + 6x + 9
In above expression
a = x, b = 3, 2ab = 2x × 3
Put values in the identity (a + b)2 = a 2+ 2ab + b2
Therefore,
x2 + 6x + 9 = (x + 3)2 Required factorisation
(b) (a – b)2 = a 2– 2ab + b2
Eg:- Factorise x2 – 6x + 9
Ans-
x2 – 6x + 9
In above expression
a = x, b = 3, 2ab = 2x × 3
Put values in the identity (a – b)2 = a 2– 2ab + b2
Therefore,
x2 – 6x + 9 = (x – 3)2 Required factorisation
(c) (a + b) (a – b) = a2– b2
Eg:- Factorise x2 – 16
Ans-
In the above expression
a2 = x2, b2 = 42
Put values in the identity (a + b) (a – b) = a2– b2
x2 – 22 = (x + 2) (x – 2) Required factorisation
Helping Topics