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

1 Answer
Jul 2, 2016

Answer:

#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)#