NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.2 in Chapter 14 - Factorisation

Question 11 Exercise 14.2

Factorise:

$9 x^{2} y^{2}-16$

Answer:

Solution:

The factorization is substantially simpler when certain identities are used.

Many expressions that need to be factored have the form of or can be put into the form of: a² - b² = ( a+b)(a-b)

$\begin{array}{l} =(3 x y)^{2}-4^{2} \\ =(3 x y-4)(3 x y+4) \\ \text { Using Identity: } x^{2}-y^{2}=(x+y)(x-y) \end{array}$

Video transcript

"hello students i am rita your master tutor and the today's question is factorize 63 a square minus 112 v square first of all we have to make these two numbers is a perfect square then so 63 and 112 both are not a perfect square of any number so first of all we find the hcf of at cf of 63 and 112 it means first of all we have to find the highest common factor of 63 and 112 so we find the hcf of 63 and 112 then it becomes 63 63 12 minus 3 9 10 minus 6 4 now 49 ones are 49 39 13 minus 9 4 5 minus 4 1 and then 14 3's are 42 9 minus two seven and then seven to the fourteen it means the highest common factor is seven so we take out seven common from these two number seven nines are sixty three a square minus seven when the seven forty two seven six of forty two now this become a perfect square because nine is a perfect square of 3 and a square is a perfect square of 8 and 16 is a perfect square of 4 now it becomes a perfect square now we use the identity of x square minus y square now we use the identity of x square minus y square and the identity of x square minus y square is x plus y into x minus y now we use this identity in this question and we got the answer that is 7 3 a minus 4 b and 3 a plus 4 b so this is the answer of this question 7 bracket 3a minus 4b and 3a plus 4v so this is the answer and we use this identity in this question i hope you like this video so please subscribe video for more updates and do comment your questions thank you "

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