How do you factor `x^2 - 8xy + 16y^2 - 3x + 12y +2`?

This is a disguised version of:

`t^2-3t+2 = (t-1)(t-2)`

with `t=x-4y` as follows:

`x^2-8xy+16y^2-3x+12y+2`

`=(x-4y)^2-3(x-4y)+2`

`=((x-4y)-1)((x-4y)-2)`

`=(x-4y-1)(x-4y-2)`

A Little Slower

This polynomial is a mixture of terms of degree `2`, `1` and `0`.

So if it factors, then it has two factors each containing a mixture of terms of degree `1` and `0`.

If we removed the terms of degree `0` from both of the factors, then the product of the simplified factors would be exactly the terms of degree `2`.

So to find the terms of degree `1` in each factor we just need to look at the terms of degree `2` in the original polynomial, namely:

`x^2-8xy+16y^2`

Note that `x^2` and `16y^2 = (4y)^2` are both perfect squares, so we might hope and indeed find that this is a perfect square trinomial:

`x^2-8xy+16y^2 = (x-4y)^2`

Next note that the terms of degree `1` in the original polynomial are a simple scalar multiple of the same `(x-4y)`, namely `-3(x-4y)`

Hence we find:

`x^2-8xy+16y^2-3x+12y+2 = (x-4y)^2-3(x-4y)+2`

Then substitute `t = (x-4y)`, to get `t^2-3t+2`, which factorises as `(t-1)(t-2)`, etc.