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)`



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