Factorize the following expressions:
4a2 + b2 + 9c2 – 4ab – 6bc + 12ca
Method 1
The above equation can be simplified as :
= (2a)2 + (– b)2 + (3c)2 + 2(2a)(– b) + 2(– b)(3c) + 2(3c)(2a)
This equation is of the form: p2 + q2 + r2 + 2pq + 2qr + 2rp
where p =2a, q = – b, r = 3c
Using the identity: p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2
We get,
⇒ 4a2 + b2 + 9c2 – 4ab – 6bc + 12ca = (2a – b + 3c)2
= (2a – b + 3c)2
Method 2
The above equation can be simplified as:
= (2a)2 + (– b)2 + (3c)2 + 2(2a)(– b) + 2(– b)(3c) + 2(3c)(2a)
= {(2a)2 + 2(2a)(– b) + (– b)2} + (3c)2 + 2(– b)(3c) + 2(3c)(2a)
(∵ p2 + 2pq + q2 = (p + q)2)
= (2a – b)2 + (3c)2 + 2(– b)(3c) + 2(3c)(2a)
Taking 2(3c) common in term 2(– b)(3c) + 2(3c)(2a)
= (2a – b)2 + (3c)2 + 2(3c)(– b + 2a)
= (2a – b)2 + 2(3c)(2a – b) + (3c)2
This of the form: (p + q)2 + 2r(p + q) + r2
Where, p = 2a, q = – b and r = 3c
Using the identity: (p + q)2 = p2 + 2pq + q2
We get,
⇒ 4a2 + b2 + 9c2 – 4ab – 6bc + 12ca = (2a – b + 3c)2
4a2 + b2 + 9c2 – 4ab – 6bc + 12ca = (2a – b + 3c)2
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