Question

using suitable identity value 101× 202​

Answer 1

Answer:

101 × 202

=> (100+1) (200+2)

=> (x+a) (x+b) = x^2+(a+b)x+ab

Putting a=1 , b=2

=> (x)^2+(1+2)x+1×2

=> x^2+3x+2

I hope this will help you

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