"Question 2 Factorise (i) 4p^2 − 9q^2 (ii) 63a^2 − 112b^2 (iii) 49x^2 − 36 (iv) 16x^5 − 144x^3 (v) (l + m)^2 − (l − m)^2 (vi) 9x^2y^2 − 16 (vii) (x^2 − 2xy + y^2) − z^2 (viii) 25a^2 − 4b^2 + 28bc − 49c^2
Class 8 Factorisation Page 223"
Answers
Factors:
In a product of two or more expressions each expression is called a factor of the product.
Factorization:
The process of writing a given expression as a product of two or more factors is called factorization.
Factorisation when a binomial is the difference of two squares:
In this case we use the formula
a² - b² = (a+b)(a-b)
Factorization when an expression is a complete square:
In this case we use the following formula
1. a²+2ab+b²= (a+b)²
2. a²-2ab+b²= (a-b)²
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Solution:
1) 4p² – 9q²
=(2p)²–(3q)²
[a²- b²=(a+b)(a-b)]
=(2p-3q)( 2p+3q)
2) 63a² – 112b²
= 7( 9a² -16b²)
=7[ (3a)² –(4b)²]
[a²- b²=(a+b)(a-b)]
=7(3a-4b)( 3a+4b)
3) 49x² – 36
=(7x)² –(6)²
[a²- b²=(a+b)(a-b)]
=(7x-6)( 7x+6)
4) 16x^5-144x³
= 16x³(x²-9)
= 16x³(x² - 3²)
[a²- b²=(a+b)(a-b)]
= 16x³(x+3)(x-3)
5) (l + m) ² – (l – m) ²
= [ l+m +l-m][l+m-l+m]
[a²- b²=(a+b)(a-b)]
=2l×2m
=4lm
6) 9x² y² – 16
(3xy)² - 4²
[a²- b²=(a+b)(a-b)]
= (3xy-4)(3xy+4)
7) (x² – 2xy + y²) – z²
= (x-y)² –z²
[ (a-b)²= a²+b² -2 ab]
[a² –b² = (a-b)(a+b)]
= (x-y+z)(x-y-z)
8) 25a² – 4b² + 28bc – 49c²
Factorizing each tem
= (5a)² –(2b)² + 2×2b×7c –(7c)²
Rearranging the terms
=(5a)²–[(2b)² - 2×2b×7c +(7c)²]
as [(a-b)²= a²+b²-2 ab]
=(5a)² –(2b-7c)²
[a²- b²=(a+b)(a-b)]
=(5a-2b+7c)(5a+2b-7c)
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Hope this help you...