2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

Rearranging as Quadratic of $x,$ $$x^2+x(2y-2)+y^2-2y-15=0$$

$$x=\frac{-(2y-2)\pm\sqrt{(2y-2)^2-4(y^2-2y-15)}}2=\frac{-(2y-2)\pm8}2=-(1-y)\pm4$$

$\endgroup$
3
$\begingroup$

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

$$c_1=-1,c_2=3$$

$\endgroup$
2
$\begingroup$

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 $x + y = -3$.

$\endgroup$
  • $\begingroup$ But equations like $$3x^2 - 2xy - y^2 - 5x + y + 2 = 0$$ , $$2x^2 - xy - 6y^2 + 3x + 22y - 20$$ , $$6x^2 + 5xy - 4y^2 + 7x + 13y - 3 = 0$$ are problems without general strategy. $\endgroup$ – user124672 Oct 7 '14 at 3:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.