Page 2 - ln

Examples 1 on 1st Identity of Factorization :
2
1) 9a + 12ab + 4b 2
2
Solution :9a + 12ab + 4b 2
2
= (3a) + 2 . (3a) . (2b) + (2b) 2
= (3a + 2b) [ since a = 3a and b = 2b; a + 2ab + b = (a + b) ]
2
2
2
2
In the 2nd identity, a - 2ab + b = (a - b) ,
2
2
2
1st and the last term should be perfect square and the middle term is two times the
square root of 1st and the last term and the sign of the middle term is negative.
Example 2 on 2nd Identity of Factorization :
2
1) 4p - 20pq + 25q 2
Solution :
2
4p - 20pq + 25q 2
2
= (2p) - 2 . (2p) . (5q) + (5q) 2
2
2
2
= (2p - 5q) [ since a = 2p and b = 5q; a + 2ab - b = (a - b) ]
2
Example 3: Find the product of (x + 1)(x + 1) using standard algebraic identities.
Solution: (x + 1)(x + 1) can be written as (x + 1) . Thus, it is of the form Identity I where a =
2
x and b = 1. So we have,

2
2
2
(x + 1) = (x) + 2(x)(1) + (1) = x + 2x + 1
2
2
Example 4: x + 10x + 25
Solution:
2
We can express x + 10x + 25 as using a + 2ab + b = (a + b) 2
2
2
2
= (x) + 2 ( x)( 5) + (5) 2
= (x + 5) 2
= (x + 5)(x + 5)
 Here 25 = is our required square to do the factorization

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