How to factor general equation of pair of straight lines with two variables and constant number at last?
How to factor the equation: $$x^2 + 2xy + y^2 - 2x - 2y - 15 = 0$$
I tried to solve this equation several times and looked for the reference about solving polynomial equations with two variables. I am confident to solve these type of equations without constant term at last. But with a constant term just like $(15)$ here, i couldn't solve it. Give it a try! and general approach to solve these type of equations.
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$\begingroup$Thinking the other way, $(a_1x+b_1y+c_1)(a_2x+b_2y+c_2)=a_1a_2x^2+(a_1b_2+b_1a_2)xy+b_1b_2y^2+(a_1c_2+c_1a_2)x+(b_1c_2+c_1b_2)y+c_1c_2$
So it's like doing factorizing quadratic formula with one variable 3 times(coefficient in front of $x^2, xy, y^2$, then $x^2,x,constant$ then $y^2, y, constant$).
The trick is to start with one subset(anyone) of coefficients then getting others by comparing coefficients, take one of your examples
$$6x^2+5xy-4y^2+7x+13y-3$$
Not hard to do the factorization $$6x^2+5xy-4y^2=(3x+4y)(2x-y)$$
Now $$6x^2+5xy-4y^2+7x+13y-3=(3x+4y+c_1)(2x-y+c_2)$$
Comparing coefficients we get
$$c_1=-1,c_2=3$$
$\endgroup$ $\begingroup$Rearranging as Quadratic of $x,$ $$x^2+x(2y-2)+y^2-2y-15=0$$
$$x=\frac{-(2y-2)\pm\sqrt{(2y-2)^2-4(y^2-2y-15)}}2=\frac{-(2y-2)\pm8}2=-(1-y)\pm4$$
$\endgroup$ $\begingroup$While I do not know a general strategy, the idea here is to notice that this is a quadratic equation in $x + y$.
\begin{align*} x^2 + 2xy + y^2 - 2x - 2y - 15 & = 0\\ (x + y)^2 - 2(x + y) - 15 & = 0\\ (x + y - 5)(x + y + 3) = 0 \end{align*} so the lines are $x + y = 5$ and $x + y = -3$.
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