The factor which is present in each term is taken as common and each term of the expression divided by the common factor and the quotient represents the factorization of the expression. For example,
ax+ay
= a(x+y)
= axy^{2} + ax^{2}y
= 2xy(y + 2x)
Factorisation of expressions having a common factor in the group of terms
In this case, the terms of the given expression are arranged in groups in such a way that each group has a common factor. For example,
a^{2} + ab + ca + bc
a(a+b) + c(a+b)
(a+b)(a+c)
Factorisation of expressions having the difference of two squared terms
The algebraic expressiona^{2} - b^{2} is the difference of two squared terms. Here a^{2} and b^{2} are the squared terms and a and b are their square roots respectively. We have learnt that a^{2} - b^{2} is the product of (a+b) and(a-b)
\(\therefore\) a^{2} - b^{2} = (a+b) (a-b)
So, (a+b) and (a-b) are the factors of a^{2} - b^{2}
To factorise such expression, we should re-write given terms in the form of a^{2} - b^{2}.
Then, a^{2} - b^{2} = (a+b) (a-b) represents the factorise. For example,
a) 4x^{2} + 9y^{2}
= (2x)^{2} - (3y)^{2}
= (2x+3y)(2x-3y)
b) 9x^{2} + 25y^{2}
= (3x)^{2} + (5y)^{2}
= (3x+5y)(3x-5y)
Factorisation of expressions of the form x^{2} + px + q
While factorising a trimonial expression of the formx^{2} + px + q, we should search any two number a and b such that a+b = p and ab = q. Clearly a and b must be the factors of q. Then px is expanded in the form ax + bx and factorization is performed by grouping.
Solution:
a^{2} + ab + ca + bc
= a(a+b) + c(a+b)
= (a+b) (a+c)
Solution:
x^{2} - 3a + 3x - ax
= x^{2}+ 3x - ax - 3a
= x(x+3) -a (x+3)
= (x+3) (x-a)
Solution:
= (9x^{2})^{2} - (4y^{2})^{2}
= (9x^{2} + 4y^{2}) (9x^{2} - 4y^{2})
= (9x^{2} + 4y^{2}) [(3x^{2})^{2} - (2y)^{2}]
= (9x^{2} + 4y^{2}) (3x + 2y) (3x - 2y)
Solution:
Here, (a+b) = 2
\(\therefore\) (a+b)^{3} = 2^{3}
or, a^{3} + 3a^{2}b + 3ab^{2} + b^{3} = 8
or, a^{3} + b^{3} + 3ab(a+b) = 8
or, a^{3} + b^{3} + 3ab×2 = 8
or, a^{3} + b^{3} + 6ab = 8
So, the reqiured value of a^{3} + b^{3} + 6ab is 8.
Solution:
= 24(40 - 20)
= 24×20
= 480
Solution:
= (65+55) (65-55)
= 120×10
= 1200
What will be factors of 2 and 3?
What will be results of 24 × 40 - 20 × 24 by simplifying?
What will be answered when simplifying 65^{2} - 55^{2}
Factorise x^{2} + 5x + 6
Factorise x^{2} - 4.
Factorise x^{2} - 3a + 3x - ax
Resolve into factors 3a(x + y) - 4b(x + y)
Simplify 79 × 81 - 5
Factorise: max - may
Factorise 36 × 24 + 26 × 36
ASK ANY QUESTION ON Factors and Factorisation
No discussion on this note yet. Be first to comment on this note