Question

solve this by factorisation method​

Answer 1

Solution :

i ) 2p² + 5pq + 2q²

= 2p² + 4pq + pq + 2q²

= 2p(p+2q)+q(p+2q)

= (p+2q)(2p+q) ---( 1 )

Now,

9x²-9(p+q)x+(2p²+5pq+2q²)=0

=> 9x²-9(p+q)x+(p+2q)(2p+q)=0

Splitting the middle term ,

we get

9x² -3(p+2q)x-3(2p+q)x

+ (p+2q)(2p+q)=0

=> 3x[3x-(p+2q)]

-(2p+q)[3x-(p+2q) ]= 0

=> [3x-(p+2q)][3x-(2p+q)]=0

Therefore ,

i ) 3x - (p+2q) = 0

=> x = (p+2q)/3

Or

ii ) 3x- (2p+q) = 0

=> x = (2p+q)/3

•••••

Answer 2

Answer:

Step-by-step explanation:

See the attachment