I was searching for a method that would allow me to factor polynomials like this one

$x^3 - 13x^2 +(14+4y)x + 8y=0$

I failed, I've only found how to factor by grouping or long division with already given root. That's useless for me in this case, what can I do here?

  • $\begingroup$ Is $y$ a function of $x$? $\endgroup$ – graydad Feb 4 '15 at 15:34
  • 1
    $\begingroup$ it's just a variable $\endgroup$ – I I Feb 4 '15 at 15:41
  • $\begingroup$ I mean, is it treated like a constant? Or could one solve for $y$ and then set $y=0$? $\endgroup$ – graydad Feb 4 '15 at 15:41
  • $\begingroup$ like a constant $\endgroup$ – I I Feb 4 '15 at 20:41

If you rearrange as $$y=-\frac{x^3-13x^2+14x}{4x+8}$$ you will see that there is precisely one real value of $y$ for each real value of $x$ except in the case $x=-2$ when there is no solution.

You can easily factorise numerator and denominator to identify zeros and asymptotes.

  • Note that the polynomial nay always not be factorizable (Irreducible Factorization) or may involve complex (not real) roots.
  • If it is possible, there may be the case:
    • If it is a cubic equation in x like the following, and if it has a rational root of the form $p/q$ then p must divide $a_3$ and q must divide $a_0$ $$a_0x^3+a_1x^2+a_2x+a_3=0$$
    • If you are really serious, you may like to see Cardano's method.
    • If you have many variables, I don't think there is a general method except noticing some pattern or some prior information.

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