Q. 5 B3.9( 24 Votes )

# Factorise9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca

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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