Factorisation of a quadratic equation

Question

I've been trying to factorise the following equation in all sorts of ways. The problem I encounter every time is not being able to do anything with the constant i.e. $-20$. Would any of you mind helping me with it? $$ 2x^2 + xy - 6y^2 - 6x + 23y - 20 = 0 $$

Answer 1

An obvious start would be $$(2x-3y)(x+2y)-6x+23y-20.$$ Now try some educated guesses to determine $a$ and $b$ from $$(2x-3y+a)(x+2y+b)=(2x-3y)(x+2y)-6x+23y-20,$$ making heavy use of $ab=-20$. You'll end up in $a=4$ and $b=-5$.

Answer 2

The general (Gauß) method consists in writing this equation as an algebraic sum of squares. If it can be factored, you obtain a difference of squares.

First group all terms with an $x$, and complete the square: \begin{align} &\phantom{{}={}}(2x^2 + xy - 6x)- 6y^2 + 23y - 20=2\bigl(x^2 +\tfrac12xy -3x\bigr)- 6y^2 + 23y - 20 \\ &=\Bigl[2\bigl(x +\tfrac14y -\tfrac32\bigr)^2-\tfrac18y^2+\tfrac32y-\tfrac92\Bigr]- 6y^2 + 23y - 20 \\ &=2\bigl(x +\tfrac14y -\tfrac32\bigr)^2-\tfrac{49}8y^2+\tfrac{49}2y-\tfrac{49}2=\tfrac18\Bigl[16\bigl(x +\tfrac14y -\tfrac32\bigr)^2-49\bigl(y^2-4y+4\bigr)\Bigr] \\ &=\tfrac18\Bigl[(4x+y-6)^2-\bigl(7(y-2)\bigr)^2\Bigr]. \end{align} Can you end the computation?

Answer 3

Note that, asymptotically, $2x^2 + xy - 6y^2=(2x-3y)(x+2y)$. Therefore, the factorized equation takes the form,

$$ 2x^2 + xy - 6y^2 - 6x + 23y - 20 = (2x-3y+k_1)(x+2y+k_2) $$

Then, setting $x=0$ in above equation,

$$6y^2-23y+20=(3y-k_1)(2y+k_2)$$

which yields $k_1=4$ and $k_2=-5$. Thus,

$$ 2x^2 + xy - 6y^2 - 6x + 23y - 20 = (2x-3y+4)(x+2y-5) $$