Factorise the following: a + 2 b 2 9 + 16 - 8 3 a + 2 b

Given, a + 2 b 2 9 + 16 - 8 3 a + 2 b Assume a + 2 b = x , then x 2 9 + 16 - 8 3 x = x 2 9 - 8 3 x + 16 Converting the expression in the form of p 2 - 2 p q + q 2 = p - q 2 = x 3 2 - 2 × x 3 × 4 + 4 2 = x 3 - 4 2 Substituting value of x , x 3 - 4 2 = a + 2 b 3 - 4 2 Hence, a + 2 b 2 9 + 16 - 8 3 a + 2 b = a + 2 b 3 - 4 2

Factorise the following: 8 1 + x 3 2 - 6 1 + x 3 - 35

Given, 8 1 + x 3 2 - 6 1 + x 3 - 35 Assume 1 + x 3 = a So, 8 1 + x 3 2 - 6 1 + x 3 - 35 = 8 a 2 - 6 a - 35 We need to find two integers such that their sum is - 6 and product is 8 × - 35 = - 28 0 . By trial and error method, we have - 20 + 14 = - 6 and - 20 × 14 = - 280 . Thus, 8 a 2 - 6 a - 35 = 8 a 2 + - 20 + 14 a - 35 = 8 a 2 - 20 a + 14 a - 35 = 4 a 2 a - 5 + 7 2 a - 5 = 4 a + 7 2 a - 5 Re-substituting value of a , we get 4 a + 7 2 a - 5 = 4 1 + x 3 + 7 2 1 + x 3 - 5 = 4 + 4 x 3 + 7 2 + 2 x 3 - 5 = 1 3 12 + 4 x + 21 1 3 6 + 2 x - 15 = 1 9 4 x + 33 2 x - 9 Hence, 8 1 + x 3 2 - 6 1 + x 3 - 35 = 1 9 4 x + 33 2 x - 9

Factorise the following: 25 x 2 + y 2 4 - 5 y x

Given, 25 x 2 + y 2 4 - 5 y x This can be written as 25 x 2 + y 2 4 - 5 y x = 25 x 2 - 5 y x + y 2 4 Convert the given expression in the form of a 2 - 2 a b + b 2 = ( a - b ) 2 = 5 x 2 - 2 × 5 x × y 2 2 + y 2 2 = 5 x - y 2 2 Hence, 25 x 2 + y 2 4 - 5 y x = 5 x - y 2 2

E M B I B E

Joy Of Mathematics 8>Chapter 13 - Factorisation>Exercise

S.N. Chhibber Solutions for Chapter: Factorisation, Exercise 3: Exercise

Author:S.N. Chhibber

S.N. Chhibber Mathematics Solutions for Exercise - S.N. Chhibber Solutions for Chapter: Factorisation, Exercise 3: Exercise

Attempt the practice questions on Chapter 13: Factorisation, Exercise 3: Exercise with hints and solutions to strengthen your understanding. Joy Of Mathematics 8 solutions are prepared by Experienced Embibe Experts.