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\ 06-Jan-2025 Bharat Arora Proof-6 Reader’s Sign _______________________ Date __________

Practice Time 13A

1. Write each of the following expressions as a product of two irreducible factors.
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(a) 6a + 12a – 18a b (b) 2mn + 3m n + 4n 3 (c) 2a b + 3b c + 6a b c
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2. Factorise each of the following by taking out their common factors.
(a) 45xy – 54x y (b) 38u – 19u (c) –3x + 3xy + 3xz
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(d) 33l mn – 44m n + 55mn 2 (e) 2x(x – y) + 3y(x – y) + 5(y – x)
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(f) –5(x – 3y) + 15(x – 3y) (g) 4(a + b)(3x – y) + 6(a + b)(2y – 3x)
3. Factorise the following by regrouping.
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(a) 2st – 3 – 6s + t (b) 3axy + 15x + 5ay + 25 (c) 1 + p + pq + p q (d) xy – x – y + 1
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(e) x – 2ax – 2ab + bx (f) 26x y + 4x y + 13x y + 8x
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Factorisation by Using Identities
We can factorise algebraic expressions by using standard identities:
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I. (x + y) = x + 2xy + y II. (x – y) = x – 2xy + y 2 III. x – y = (x + y)(x – y)
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To factorise an expression, start by identifying any common factors in all terms and factoring
them out. Then, arrange the remaining terms in either ascending or descending order of the
variable’s powers and then check for the following conditions:
1. The first and last terms should be perfect squares of some monomial factors.
2. The middle term should equal to twice the product of those monomial factors (whose squares
make up the first and last terms).
Let us consider examples based on the above identities.
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Example 4: Factorise x + 6x + 9. Get it right!
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Solution: x + 6x + 9 = (x) + 2(x)(3) + (3) = (x + 3) 2 l (3x + 7) = 9x + 49
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= (x + 3)(x + 3) [Using Identity I] l (3x + 7) = (3x) + (7) + 2(3x)(7)
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Example 5: Factorise x – 18xy + 81y . = 9x + 49 + 42x
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Solution: x – 18xy + 81y = (x) – 2(x)(9y) + (9y) = (x – 9y) 2
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= (x – 9y)(x – 9y) [Using Identity II]
Example 6: Factorise:
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(a) x – 169 (b) 4p – 36
Solution: (a) x – 169 = x – (13) = (x + 13)(x – 13) [Using Identity III]
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(b) 4p – 36 = 4(p – 9) = 4(p – 3 ) = 4(p + 3)(p – 3) [Using Identity III]
Example 7: Factorise 4x + 12xy + 9y – z 2
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Solution: (4x + 12xy + 9y ) – z = [(2x) + 2(2x)(3y) + (3y) ] – z 2 Quick Check
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= (2x + 3y) – z [Since, (x + y) = x + 2xy + y ] Fill in the blanks:
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= (2x + 3y + z)(2x + 3y – z) [Since, (x + y)(x – y) = x – y ] 1. 100 – 49p = ...............
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2. (105) – (95) = ...............
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Thus, 4x + 12xy + 9y – z = (2x + 3y + z)(2x + 3y – z)
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Mathematics-8 302