# Question #83abc

Oct 19, 2016

No useful factorization for $f \left(x , y\right) = {x}^{2} + 6 x y + 9 {y}^{2} - 21 y + 12$

#### Explanation:

$f \left(x , y\right) = {x}^{2} + 6 x y + 9 {y}^{2} - 21 y + 12$

If $f \left(x , y\right)$ is factorable, that implies in multiple tracing of $f \left(x , y\right) = 0$

So, if $f \left(x , y\right)$ is factorable for instance into ${f}_{1} \left(x , y\right) {f}_{2} \left(x , y\right)$ this means that

$f \left(x , y\right) = {f}_{1} \left(x , y\right) {f}_{2} \left(x , y\right) = 0$ has a minimum of two null leafs. One for ${f}_{1} \left(x , y\right) = 0$ and the other for ${f}_{2} \left(x , y\right) = 0$

In our case $f \left(x , y\right)$ represents a slanted parabola having one null leaf, so this $f \left(x , y\right)$ is not factorable.

We know that $f \left(x , y\right)$ represents a parabola because putting it in the form

$f \left(x , y\right) = \frac{1}{2} \left(x , y\right) \cdot M \cdot \left(\begin{matrix}x \\ y\end{matrix}\right) + \left(x , y\right) \cdot b + c$

$M = \left(\begin{matrix}2 & 6 \\ 6 & 18\end{matrix}\right)$ has characteristic polynomial

${p}_{M} \left(s\right) = {s}^{2} - 20 s = s \left(s - 20\right)$

which has a null root.

Attached the null trace of $f \left(x , y\right) = 0$