Class 9 Maths


Polynomials

Exercise 2.5 Part 4

Question 4: Expand each of the following using suitable identities:

(i) `(x + 2y + 4z)^2`

Answer: Given, `(x + 2y + z)^2`

Let, a = x, b = 2y and c = 4z

We know that, `(a + b + c)^2`

`= a^2 + b^2 + c^2 + 2ab + 2bc + 2ac`

So, the given expression can be written as follows:

`x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz`

(ii) `(2x – y + z)^2`

Answer: Given, `(2y – y + z)^2`

`= [2x + (-y) +z]^2`

`= Let, `a = 2x`, `b = - y` and `c = z`

Using the identity
`(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc +2ac`,
we get: `(2x)^2 + (-y)^2 + z^2 + 2(2x(-y) + 2(-y)z + 2(2xz)`

`= 4x^2 + y^2 + z^2 + 2(-2xy) + 2(-yz) + 4xz`
`= 4x2 + y2 + z2 - 4xy – 2yz + 4xz`

(iii) `( -2x + 3y + 2z)^2`

Answer: Given, `(- 2x + 3y + 2z)^2`
Let, `a= - 2x`, `b = 3y` and `c = 2z`

Using the identity
`(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac`
we get:`(-2x)^2 + (3y)^2 + (2z)^2 + 2( - 2x\xx3y) + 2(3y\xx2z) + 2(-2x\xx2z)`

`= 4x^2 + 9y^2 + 4z^2 + 2( - 6xy) + 2(6yz) + 2(-4xz)`

`= 4x^2 + 9y^2 + 4z^2 – 12xy + 12yz – 8xz`