Subject: Compulsory Maths

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.

- The terms of the given expression are arranged in groups in such a way that each group has a common factor.
- The algebraic expression a
^{2}- b^{2}is the difference of two squared terms.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Factorise: ax + ay

Solution:

ax + ay

= a(x+y)

Factorise: 2px^{2} - 6p^{2}x

Solution:

2px×x -2×3p×px

= 2px(x-3p)

Resolve into factors: 3a(x+y) - 4b(x+y)

Solution:

3a(x+y) - 4b(x+y)

(x+y) (3a - 4b)

a^{2} + ab + ca + bc

Solution:

a^{2} + ab + ca + bc

= a(a+b) + c(a+b)

= (a+b) (a+c)

Factorise: x^{2} - 3a + 3x - ax

Solution:

x^{2} - 3a + 3x - ax

= x^{2}+ 3x - ax - 3a

= x(x+3) -a (x+3)

= (x+3) (x-a)

Factorise: 81x^{4} - 16y^{4}

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)

If (a+b) = 2, find the value of a3 + b3 + 6ab.

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.

Simplify: 24×40 - 20×24

Solution:

= 24(40 - 20)

= 24×20

= 480

Simplify by factorization process: 65^{2} - 55^{2}

Solution:

= (65+55) (65-55)

= 120×10

= 1200

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