Question

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

Answer 1

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)`