How do you factor #4x^2 - 6xy + 9y^2#?

1 Answer
Aug 4, 2016

Answer:

Factors of #4x^2-6xy+9y^2# are #(4x-3y-3sqrt3yi)(x-3/4y+3/4sqrt3yi)#

Explanation:

#4x^2-6xy+9y^2# is a homogeneous quadratic function and factorizing it similar to those for quadratic function such as #ax^2+bx+c#, as taking #y^2# out, makes it #y^2(4x^2/y^2-(6xy)/y^2+9)=y^2(4(x/y)^2-6x/y+9)=y^2(4u^2-6u+9)#, where #u=x/y#.

As the discriminant #b^2-4ac=36-144=-108# is negative, we can only have complex factors.

Now zeros of #4x^2-6xy+9y^2# are #(-(-6)+-sqrt((-6)^2-4xx4xx9))/(2xx4)# or #(6+-sqrt(-108))/8# or #3/4+-3/4sqrt3i# and factors of #4u^2-6u+9# could be

#4(u-3/4-3/4sqrt3i)(u-3/4+3/4sqrt3i)# or #(4u-3-3sqrt3i)(u-3/4+3/4sqrt3i)#

Hence factors of #4x^2-6xy+9y^2# are #(4x-3y-3sqrt3yi)(x-3/4y+3/4sqrt3yi)#