Question

2x^2y+2x^2y^3-3x^3y^2-3xy^4 factorise​

Answer 1

Answer:

Given : x

3

−2x

2

y+3xy

2

−6y

3

By taking x

2

as common in the first two term and 3y

2

as common in the second two term

x

3

−2x

2

y+3xy

2

−6y

3

=x

2

(x−2y)+3y

2

(x−2y)

So we get,

x

3

−2x

2

y+3xy

2

−6y

3

=(x−2y)(x

2

+3y

2

)