Notes on Factors and Factorisation | Grade 7 > Compulsory Maths > Factorisation, H.C.F. and L.C.M. | KULLABS.COM

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#### Factorisation of expressions which have a common factor in each of its terms

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)

= axy2 + ax2y

= 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,

a2 + 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 expressiona2 - b2 is the difference of two squared terms. Here a2 and b2 are the squared terms and a and b are their square roots respectively. We have learnt that a2 - b2 is the product of (a+b) and(a-b)

$$\therefore$$ a2 - b2 = (a+b) (a-b)

So, (a+b) and (a-b) are the factors of a2 - b2

To factorise such expression, we should re-write given terms in the form of a2 - b2.

Then, a2 - b2 = (a+b) (a-b) represents the factorise. For example,

a) 4x2 + 9y2

= (2x)2 - (3y)2

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

b) 9x2 + 25y2

= (3x)2 + (5y)2

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

Factorisation of expressions of the form x2 + px + q

While factorising a trimonial expression of the formx2 + 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 a2 - b2 is the difference of two squared terms.
.

### Very Short Questions

Solution:

a2 + ab + ca + bc

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

= (a+b) (a+c)

Solution:

x2 - 3a + 3x - ax

= x2+ 3x - ax - 3a

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

= (x+3) (x-a)

Solution:

= (9x2)2 - (4y2)2

= (9x2 + 4y2) (9x2 - 4y2)

= (9x2 + 4y2) [(3x2)2 - (2y)2]

= (9x2 + 4y2) (3x + 2y) (3x - 2y)

Solution:

Here, (a+b) = 2

$$\therefore$$ (a+b)3 = 23

or, a3 + 3a2b + 3ab2 + b3 = 8

or, a3 + b3 + 3ab(a+b) = 8

or, a3 + b3 + 3ab×2 = 8

or, a3 + b3 + 6ab = 8

So, the reqiured value of a3 + b3 + 6ab is 8.

Solution:

= 24(40 - 20)

= 24×20

= 480

Solution:

= (65+55) (65-55)

= 120×10

= 1200

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2
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430
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480

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• ### Factorise x2 + 5x + 6

(x - 2) (x + 3)
(x + 2) (x + 3)
(x - 2) (x - 3)
(x + 3) (x + 2)
• ### Factorise x2 - 4.

(x + 1) (x - 4)
(x - 2) (x + 2)
(x + 2) (x - 1)
(x + 2) (x - 2)
• ### Factorise x2 - 3a + 3x - ax

(x - 2) (x + a)
(x - a) ( x + a)
(x + 1) (x + a)
(x + 3) ( x - a)
• ### Resolve into factors 3a(x + y) - 4b(x + y)

(x + y) (a + b)
(x + y) (3a - 4b)
(x - y) (2a + 2b)
(x + y) (a - b)

6395
6399
6391
6394

m(x - y + a)
ma(x - y)
mx - xy
am(x + y)

1600
1800
1500
1700

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