Question

find the product using algebraic identity.(99*103)

Answer 1

hey there

we can write 99 and 103 as

(100-1) (100+3)
using identity IV (x+a) (x+b) = x^2+(a+b)x+ab
(100)^2 + (-1+3)100+(-1)×(3)
10000+200-3
=10198..

HOPE THIS HELPS. .........

Answer 2

(100-1)(100-(-3))
(100)^2 - 100(1+(-3)) + (-3)
10000 - (-200) - 3
10197 is the answer.
[(x-a)(x-b)=x^2 - x(a+b) + ab]

Answer 3

(100-1)(100-(-3))
(100)^2 - 100(1+(-3)) + (-3)
10000 - (-200) - 3
10197 is the answer.
[(x-a)(x-b)=x^2 - x(a+b) + ab]