2. find out the h. c.f
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2a , 1st exp= a2 + ab

= a(a+b)

2nd exp = a2-b2

= (a+b) (a-b)

HCF = (a+b)

b, 1 st exp= (a+b)2

= (a+b)(a+)

2 nd exp = a2-b2

= (a+b) ( a-b)

HCF = (a+b)

c, 1 st exp= (a-1)2

= (a+1) (a-1)

2nd exp= a2-1

= (a+1)(a-1)

HCF = (a-1)

d, 1st exp= (x-2)(x-3)

= x(x-3)-2(x-3)

= x2 - 3x - 2x + 6

= x2- 5x+ 6

= x2 - (3+2)x + 6

=x2 -3x-2x+ 6

= x(x-3) -2(x-3)

=(x-3)(x-2)

2 nd exp = (x+2) (x-3)

= x( x-3) + 2 (x-3)

= x2 - 3 x+ 2x -6

= x(x-3) + 2(x-3)

=(x-3) ( x+ 2)

HCF = (x-3)

3

step-by-step explanation: