Question

How do you factor `9x^2 - 18xy + 5y^2`?

Answer 1

`9x^2-18xy+5y^2 = (3x-5y)(3x-y)`

To find this, suppose `(9x^2-18xy+5y^2)=(ax+by)(cx+dy)`

Then from multiplying out the right hand side and comparing coefficients we find:

`9 = ac`

`-18 = ad+bc`

`5=bd`

We might as well choose `a` and `c` to both be positive.

`bd=5` tells us that either both `b` and `d` are positive or `b` and `d` are both negative.

Since we have chosen `a` and `c` to be positive, we can deduce from the `-18 = ad+bc` equation that `b` and `d` are both negative.

So `bd = 5` only factors as `-5 xx -1` or `-1 xx -5`.

As everything so far is symmetric, let's choose `b=-5` and `d=-1`

`ac=9` will only factor as `1xx9`, `3xx3` or `9xx1`.

Trying each of these possibilities, we find that `-18=ad+bc` is satisfied when `a=3` and `c=3`.