Question

factorise a: (x+1)^2-(y-1)^2 b: 16x^4- 81y^4 q2: factorise (a+b)^2-14c(a+b)+49c^2

Answer 1

Hi ...dear..
.
here is your answer....
in part a and b I used formula of a²-b² =(a+b)(a-b)..
.

in second question
let the a+b be x so it will become whole square of (x-7c)...
put the value of x in last......see picture

hope it helped you.....
Regards.
#shubhendu

Answer 2

(a) 

Given Equation is (x + 1)^2 - (y - 1)^2

We know that a + b)^2 = a^2 + b^2 + 2ab, (a - b)^2 = a^2 + b^2 - 2ab.

= > x^2 + 1 + 2x - (y^2 + 1 - 2y)

= > x^2 + 1 + 2x - y^2 - 1 + 2y

= > x^2 + 2x + 2y - y^2

= > x^2 - y^2 + 2x + 2y

We know that a^2 - b^2 = (a + b)(a - b)

= > (x + y)(x - y) + 2(x + y)

= > (x + y)[x - y+ 2].


Therefore factorization of (x + 1)^2 - (y - 1)^2 = (x + y[x - y + 2]



(2)

Given 16x^4 - 81y^4

= > (4x^2)^2 - (9y^2)^2

= > (4x^2 + 9y^2)(4x^2 - 9y^2)

= > (4x^2 + 9y^2)((2x)^2 - (3y)^2)

= > (4x^2 + 9y^2)((2x + 3y)(2x - 3y))

= > (4x^2 + 9y^2)(2x + 3y)(2x - 3y)


Therefore factorization of 16x^4 - 81y^4 = (4x^2 + 9y^2)(2x + 3y)(2x - 3y).



(3)

(a + b)^2 - 14c(a + b) + 49c^2

= > (a + b)^2 - 7c(a + b) - 7c(a + b) + 49c^2

= > (a + b)[a + b - 7c] - 7c[a + b - 7c]

= > (a + b - 7c)(a + b - 7c)

= > (a + b - 7c)^2.



Therefore Factorization of (a + b)^2 - 14c(a + b) + 49c^2 = (a + b - 7c)^2.




Hope this helps!