# Factorize 2 x^2 + y^2 + 3 x y + 6 x - 15 = 0 ?

Jul 27, 2016

$\left(x + \frac{y}{2} - 3\right) \left(2 x + 2 y + 12\right) + 21$

#### Explanation:

Here $f \left(x , y\right) = 2 {x}^{2} + {y}^{2} + 3 x y + 6 x - 15$

First we will classify the kind of conic we are handling. To do that we compute the matrix

$H = \left(\begin{matrix}{f}_{x x} & {f}_{x y} \\ {f}_{y x} & {f}_{y y}\end{matrix}\right)$

where ${f}_{{x}_{1} {x}_{2}} = \frac{{\partial}^{2} f}{\partial {x}_{1} \partial {x}_{2}}$

obtaining

$H = \left(\begin{matrix}4 & 3 \\ 3 & 2\end{matrix}\right)$ with characteristic polynomial

$p \left(\lambda\right) = {\lambda}^{2} - 6 \lambda - 1$

whose roots are $\left\{3 - \sqrt{10} , 3 + \sqrt{10}\right\}$

being real and with opposite singns, characterizing a hyperbola with structure

$\frac{1}{2} \left(p - {p}_{0}\right) . H . \left(p - {p}_{0}\right) + c$

Now, solving for ${p}_{0} , c$ the condition

$\frac{1}{2} \left(p - {p}_{0}\right) . H . \left(p - {p}_{0}\right) + c - f \left(x , y\right) = 0 , \forall \left\{x , y\right\} \in {\mathbb{R}}^{2}$

 { (2 x_0^2 + 3 x_0 y_0 + y_0^2 + 15 + c = 0), ( 3 x_0 + 2 y_0 =0), (4 x_0 + 3 y_0 +6= 0) :}

we obtain ${x}_{0} = 12 , {y}_{0} = - 18 , c = 21$

A possible factorization is

$\frac{1}{2} \left(p - {p}_{0}\right) . H . \left(p - {p}_{0}\right) = \left({c}_{1} x + {c}_{2} y + {c}_{3}\right) \left({c}_{4} x + {c}_{5} y + {c}_{6}\right)$

giving

$f \left(x , y\right) = \left(x + \frac{y}{2} - 3\right) \left(2 x + 2 y + 12\right) + 21$

Attached a plot with the hyperbola and the factorized lines.