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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2$\begingroup$ are you sure that there isn't a typo? $\endgroup$– Dr. Sonnhard GraubnerFeb 18, 2018 at 16:26
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$\begingroup$ Solve for $x$ or $y$ $\endgroup$– lab bhattacharjeeFeb 18, 2018 at 16:32
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$\begingroup$ It only says factor $\endgroup$– JanjanFeb 18, 2018 at 16:40
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3 Answers
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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.
answered Feb 18, 2018 at 16:48
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$\begingroup$ @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. $\endgroup$ 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 $$\left(y-\frac{3}{4}x-2-\frac{1}{4}\sqrt{x^2+40x+112}\right)\left(y+\frac{3}{4}x+2-\frac{1}{4}\sqrt{x^2+40x+112}\right)=0$$
answered Feb 18, 2018 at 16:33
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$\begingroup$ But there is no equation in the question. $\endgroup$ Feb 18, 2018 at 16:43
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$\begingroup$ but i made this to an equation $\endgroup$ Feb 18, 2018 at 16:44
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$\begingroup$ 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.
