POLYNOMIALS 45
Example 17 : Evaluate 105 × 106 without multiplying directly.
Solution : 105 × 106 = (100 + 5) × (100 + 6)
= (100)^2 + (5 + 6) (100) + (5 × 6), using Identity IV
= 10000 + 1100 + 30
= 11130You have seen some uses of the identities listed above in finding the product of some
given expressions. These identities are useful in factorisation of algebraic expressions
also, as you can see in the following examples.
Example 18 : Factorise:
(i) 49a^2 + 70ab + 25b^2 (ii)2
(^252)
49
x y
Solution : (i) Here you can see that
49 a^2 =(7a)^2 , 25b^2 = (5b)^2 , 70ab = 2(7a) (5b)
Comparing the given expression with x^2 + 2xy + y^2 , we observe that x = 7a and y = 5b.
Using Identity I, we get
49 a^2 + 70ab + 25b^2 =(7a + 5b)^2 = (7a + 5b) (7a + 5b)
(ii) We have
2 22
(^252) –– 5
492 3
yy
xx✄✁☎ ✂✆ ✁☎ ✂✆
✝ ✞ ✝ ✞
Now comparing it with Identity III, we get
2
(^252) –
49
y
x =
22
5
–
23
y
✁☎ x✂✆ ✁☎ ✂✆
✝ ✞ ✝ ✞=
55
2323
yy
✟☞ xx✡ ✠✟✌☞ ☛ ✠✌
✍ ✎✍ ✎So far, all our identities involved products of binomials. Let us now extend the Identity
I to a trinomial x + y + z. We shall compute (x + y + z)^2 by using Identity I.
Let x + y = t. Then,
(x + y + z)^2 =(t + z)^2
=t^2 + 2tz + t^2 (Using Identity I)
=(x + y)^2 + 2(x + y)z + z^2 (Substituting the value of t)