Q. 5 B3.9( 24 Votes )

Factorise

9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca

Answer :

9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca


9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = 9a2 + 4b2 + 16c2 + 12ab + (-16bc) + (-24ca)


9a2 can be written as (3a)2


4b2 can be written as (2b)2


16c2 can be written as (-4c)2


12ab can be written as 2(3a)(2b)


-16bc can be written as 2(2b)(-4c)


-24ca can be written as 2(-4c)(3a)


9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = (3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) …(i)


Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


Comparing (3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) with x2 + y2 + z2 + 2xy + 2yz + 2zx we get


x = 3a, y = 2b and z = -4c


therefore


(3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) = (3a + 2b + (-4c))2


From (i)


9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = (3a + 2b – 4c)2


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