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

May 11, 2016

${x}^{2} - 8 x y + 16 {y}^{2} - 3 x + 12 y + 2 = \left(x - 4 y - 1\right) \left(x - 4 y - 2\right)$

#### Explanation:

This is a disguised version of:

${t}^{2} - 3 t + 2 = \left(t - 1\right) \left(t - 2\right)$

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

${x}^{2} - 8 x y + 16 {y}^{2} - 3 x + 12 y + 2$

$= {\left(x - 4 y\right)}^{2} - 3 \left(x - 4 y\right) + 2$

$= \left(\left(x - 4 y\right) - 1\right) \left(\left(x - 4 y\right) - 2\right)$

$= \left(x - 4 y - 1\right) \left(x - 4 y - 2\right)$

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:

${x}^{2} - 8 x y + 16 {y}^{2}$

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

${x}^{2} - 8 x y + 16 {y}^{2} = {\left(x - 4 y\right)}^{2}$

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

Hence we find:

${x}^{2} - 8 x y + 16 {y}^{2} - 3 x + 12 y + 2 = {\left(x - 4 y\right)}^{2} - 3 \left(x - 4 y\right) + 2$

Then substitute $t = \left(x - 4 y\right)$, to get ${t}^{2} - 3 t + 2$, which factorises as $\left(t - 1\right) \left(t - 2\right)$, etc.