Class 9 Maths

# Polynomials

## Exercise 2.5 Part 9

Question: 8 – Factorise each of the following:

(i) 8a^3 + b^3 + 12a^2\b + 6ab^2

Answer: Given; 8a^3 + b^3 + 12a^2\b + 6ab^2

= (2a)^3 + b^3 + 3(2a)^2\b + 3(2ab^2)

Let x = 2a and y = b

Using the identity (x + y)^3= x^3 + y^3 + 3x^2\y + 3xy^2

We get; (2a + b)^3 = (2a + b)(2a + b)(2a + b)

(ii) 8a^3 - b^3 - 12a^2\b + 6ab^2

Answer: Given; 8a^3 - b^3 - 12a^2\b + 6ab^2

= (2a)^3 - b^3 - 3(2a)^2\b + 3(2a)b^2

Let x = 2a and y = b

Using the identity (x – y)^3= x^3 - y^3 - 3x^2\y + 3xy^2

We get; (2a – b)^3 = (2a – b)(2a - b)(2a – b)

(iii) 27 – 125a^3 - 135a + 225a^2

Answer: Given; 27 – 125a^3 - 135a + 225a^2

= 3^3 - (5a)^3 - 3 xx 3^2\ xx 5a + 3 xx (5a)^2\ xx 3

Let, x = 3 and y = 5a

Using the identity (x – y)^3= x^3 - y^3 - 3x^2\y + 3xy^2

We get; (3 – 5a)^3= (3 – 5a)(3 – 5a)(3 – 5a)

(iv) 64a^3 - 27b^3 - 144a^2\b + 108ab^2

Answer: Given: 64a^3 - 27b^3 - 144a^2\b + 108ab^2

= (4a)^3 - (3b)^3 - 3 xx (4a)^2\b + 3 xx a xx (3b)^2

Let, x = 4a and y = 3b

Using the identity (x – y)^3= x^3 - y^3 - 3x^2\y + 3xy^2

We get; (4a – 3b)^3= (4a – 3b)(4a – 3b)(4a – 3b)

(v) 27p^3-(1)/(216)-9/2\p^2+1/4\p

Solution: Given, 27p^3-(1)/(216)-9/2\p^2+1/4\p

=(3p)^3-(1/6)^3-3(3p)^2\xx1/6+3xx3pxx(1/6)^2

Using identity (x-y)^3=x^3-y^3-3x^2\y+3xy^2

We get: (3p-1/6)^3=(3p-1/6)(3p-1/6)(3p-1/6)