Q. 3 B5.0( 1 Vote )

# Factorize the following expressions:a2 + 4b2 + 36 – 4ab + 24b – 12a

= a2 + 4b2 + 36 – 4ab + 24b – 12a

Method 1

The above equation can be simplified as :
= (– a)2 + (2b)2 + (6)2 + 2(– a)(2b) + 2(2b)(6) + 2(– a)(6)
This equation is of the form: p2 + q2 + r2 + 2pq + 2qr + 2rp

where p = – a, q = 2b, r = 6
Using the identity: p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2
We get,
a2 + 4b2 + 36 – 4ab + 24b – 12a = (– a + 2b + 6)2

= (– a + 2b + 6)2

= (– 1)2(a – 2b – 6)2

= (a – 2b – 6)2

Method 2

The above equation can be simplified as

= (– a)2 + (2b)2 + (6)2 + 2(– a)(2b) + 2(2b)(6) + 2(– a)(6)
= {(– a)2 + 2(– a)(2b) + (2b)2} + (6)2 + 2(2b)(6) + 2(– a)(6)
(
a2 + 2ab + b2 = (a + b)2)
(– a + 2b)2 + (6)2 + 2(2b)(6) + 2(– a)(6)
Taking 2(6) common in term 2(2b)(6) + 2(– a)(6)
= (– a + 2b)2 + 2(6)(– a + 2b) + (6)2

This of the form: (p + q)2 + 2r(p + q) + r2

Where, p = – a, q = 2b and r =6

Using the identity: (p + q)2 = p2 + 2pq + q2

We get,

a2 + 4b2 + 36 – 4ab + 24b – 12a = (– a + 2b + 6)2

= (– a + 2b + 6)2

= (– 1)2(a – 2b – 6)2

= (a – 2b – 6)2

a2 + 4b2 + 36 – 4ab + 24b – 12a = (– a + 2b + 6)2

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