Page 90 - Maths Skills - 8

88 Maths

Exercise 6.1

1. Factorise the following algebraic expressions:

(i) 6a + 24 (ii) 8b – 4b (iii) 4x + 8x + 12
2
2
(iv) 15ax – 6ay (v) 15xy – 25x y + 10xy (vi) 3a b c + 27a b c – 81abc
3 3 2
2
2 2 2
2
(vii) 9x – 15x – 18x (viii) 36y – 18y + 54y 4
4
2
2
3
2. Factorise the following:
(i) 8m – 12m + 10m – 15 (ii) m – 4m + 3m – 12 (iii) 2t – 4t + t – 2
2
2
2
(iv) 6y – 15y + 4y – 10 (v) ab – bc – ab + c 2 (vi) a – 3a + a – 3
2
2
2
2
3
(vii) 10xz – 6zx + 8yz – 5zy (viii) 21x b – 6x b – 21xb + 6b
2
3
3. Factorise:
(i) 12uv + 3uk + 8kv + 2k (ii) 6xy – 15ax + 18x – 5ay
2
2
(iii) 192xy + 32x + 24my + 4m (iv) 112m n + 14m + 56bm n + 7bm 13
10
14
11
(v) 84xy + 196x – 36y – 84 (vi) ab + (a – 1) b – 1
2
FACTORISATION BY USING STANDARD IDENTITIES
We have learned some standard identities in the previous chapter. Those identities can be used to factorise the
algebraic expressions. In order to factorise perfect square trinomials, we may use the following identities:
a + 2ab + b = (a + b) = (a + b) (a + b)
2
2
2
a – 2ab + b = (a – b) = (a – b) (a – b)
2
2
2
Another identity that can be used to factorise the algebraic expressions is:
(a – b ) = (a + b) (a – b)
2
2
Let us learn through examples.
Let’s Attempt
Example 1: Factorise: 9a + 30a + 25 Example 2: Factorise: x – 12x + 36
2
2
Solution: 9a + 30a + 25 = (3a) + 2 × 3a × 5 + 5 2 Solution: x – 12x + 36 = x – 2 × x × 6 + 6 2
2
2
2
2
= (3a + 5) 2 = (x – 6) 2
= (x – 6) (x – 6)
= (3a + 5) (3a + 5)
Example 4: Factorise: b + b + 1
2
Example 3: Factorise: 1 – 6r + 9r 2 4 2
2
2
Solution: 1 – 6r + 9r = 1 – 2 × 1 × 3r + (3r) 2 Solution: b + b + 1 = b + 2 × b × 1 + 1
2
2

2
4
2
2
[Since, 1 = 1] 2

= b 1
= (1 – 3r) 2
2

= (1 – 3r) (1 – 3r)
= b 1 b 1

2 2