Question

How do you factor `4x^2+27xy+18y^2`?

Answer 1

`4x^2+27xy+18y^2=(4x+3y)(x+6y)`

Explanation:

To factorize such a homogeneous (as degree of varibles is two in the poynomial) quadratic algebraic expression,

one needs to break middle term `27` so that their product is equal to product of coefficients of other two terms i.e. `4xx18=72`. It is apparent that these are `24` and `3`. Hence,

`4x^2+27xy+18y^2`

= `4x^2+24xy+3xy+18y^2`

= `4x(x+6y)+3y(x+6y)`

• `(4x+3y)(x+6y)`