Question

factorise 9a2+4b2+16c2+12ab_16bc_24ca​

Answer 1

$\bf \huge </p><p>Answer$

9a² + 4b²+ 16c² + 12ab - 16bc - 24ca

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

9a²can 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