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Factor $x^2 - 3xy + 2y^2 + x -8y - 6$

Attempt at a solution:

I have factored these and don't know how to continue...

$x^2-3xy +2y^2 = (x - y) (x-2y)$

$x^2 + x -6 = (x + 3) (x - 2)$

$2y^2 - 8y + 6 = 2 (y - 3)(y - 1)$

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    $\begingroup$ are you sure that there isn't a typo? $\endgroup$ Feb 18, 2018 at 16:26
  • $\begingroup$ Solve for $x$ or $y$ $\endgroup$ Feb 18, 2018 at 16:32
  • $\begingroup$ It only says factor $\endgroup$
    – Janjan
    Feb 18, 2018 at 16:40

3 Answers 3

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Look for in form: $$x^2-3xy+2y^2+x-8y-6=(x+Ay+B)(x+Cy+D) $$ Plug $y=0$: $$x^2+x-6=(x+B)(x+D) \Rightarrow B=3; D=-2.$$ Plug $x=0$: $$2y^2-8y-6=(Ay+3)(Cy-2) \Rightarrow \begin{cases} AC=2 \\ -2A+3C=-8 \end{cases}$$ Can you finish?

Appendix: Note that the found parameters will not be suitable. So this method may not always work.

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  • $\begingroup$ Is this a formula or something? $\endgroup$
    – Janjan
    Feb 18, 2018 at 16:58
  • $\begingroup$ I would call it a method... $\endgroup$
    – farruhota
    Feb 18, 2018 at 17:02
  • $\begingroup$ @orion, yes, you are right, all but $xy$ term is not matching. So now the OP asker can turn to Dr.SonnhardGraubner's method of expressing as a quadratic equation and solving by discriminant. Thank you for pointing to the issue. I will leave my answer as an example of failed method. $\endgroup$
    – farruhota
    Feb 18, 2018 at 17:52
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you can write your equation in the form $$y^2-y\left(4+\frac{3}{2}x\right)+\frac{x^2+x-6}{2}=0$$ and solve this for $y$ you will get $$\left(y-\frac{3}{4}x-2-\frac{1}{4}\sqrt{x^2+40x+112}\right)\left(y+\frac{3}{4}x+2-\frac{1}{4}\sqrt{x^2+40x+112}\right)=0$$

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  • $\begingroup$ What equation do you mean? $\endgroup$ Feb 18, 2018 at 16:41
  • $\begingroup$ $$x^2-3xy+2y^2+x-8y-6=0$$ $\endgroup$ Feb 18, 2018 at 16:42
  • $\begingroup$ But there is no equation in the question. $\endgroup$ Feb 18, 2018 at 16:43
  • $\begingroup$ but i made this to an equation $\endgroup$ Feb 18, 2018 at 16:44
  • $\begingroup$ and by the way there are three equations $\endgroup$ Feb 18, 2018 at 16:45
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Since the polynomial is of degree $2$, we can use the well established "tool-set" for the study of Quadrics or Conic Sections.
So $$ \eqalign{ & Q(x,y) = x^{\,2} - 3xy + 2y^{\,2} + x - 8y - 6 = \cr & = \left( {x,y,1} \right)^T \left( {\matrix{ 1 & { - 3/2} & {1/2} \cr { - 3/2} & 2 & { - 4} \cr {1/2} & { - 4} & 6 \cr } } \right)\left( {\matrix{ x \cr y \cr 1 \cr } } \right) \cr} $$ But the determinant of the matrix defining the Conic $$ {\bf A}_{\,Q} = \left( {\matrix{ 1 & { - 3/2} & {1/2} \cr { - 3/2} & 2 & { - 4} \cr {1/2} & { - 4} & 6 \cr } } \right) $$ is not null, which means that the conic is not degenerate and thus $Q(x,y)$ cannot be factored, not even in the complex field.

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