Question

Factorise 112p^2-7q^2 using identity

Answer 1

Solution -

112p² - 7q²

= 7[16p² - q²]

= 7[ (4p)²- (q)² ]

= 7[ (4p + q) (4p - q) ]

= 7 (4p + q) (4p - q)

Answer - 7(4p + q) (4p - q)

Identity Used -

a² - b² = (a + b) (a - b)

Where,

a² = (4p)²

b² = (q)²

Answer 2

Answer :

To factorise

112p^2 -7q^2

Solution :

112p^2 - 7q^2

Taking 7 as common from both the terms of the algebraic expression. We get :-

7( 16p^2 - q^2)

Now, we can see that 16p^2-q^2 is in the form of a^2 - b^2.

This is a identity.

The identity states that :-

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

In this case , a = 4p and b = q

Substituting the values in the identity :-

7(4p + q )(4p - q)  [ This is the factorised form ]

So,

7 , (4p+q),and (4p-q) are factors of 112p^2 - 7q^2