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

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Answer 1 · 2016-05-11T18:52:58

$x^2-8xy+16y^2-3x+12y+2=(x-4y-1)(x-4y-2)$

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.

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