Question 86 (xii)
Using suitable identities, evaluate the following:
98×103
We have, 98×103=(100−2)(100+3)=(100)2+(−2+3)100+(−2)×3=10000+100−6 =10094 [using the identity, (x+a)(x+b)=x2+(a+b)x+ab]
Question 86 (xi)
101×103
Question 86 (iii)
(103)2