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Ok so I can factor easily regular quadratic polynomials, i.e. $5x^2+7x+9$ (I'm not sure whether that's prime, just made it up), and I was working on solving $y^2+(x^2+2x−2)y+(x^3−x^2−2x)$ by distributing the $y$ after the first parentheses and now I'm stuck. Am I just missing some skill in factoring multivariate polynomials? Don't tell me stuff about rings and fields because that won't help my situation. Thanks

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  • $\begingroup$ You may wish to have a look at our basic MathJax tutorial to learn how to typeset math here. $\endgroup$ – A.P. Mar 26 '15 at 20:55
  • $\begingroup$ For multivariate polynomials you need to define a "monomial order" to try and factor them. $\endgroup$ – mathreadler Mar 26 '15 at 20:56
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    $\begingroup$ As you noticed, factoring polynomials in more than one variable is considerably harder. Isolating one variable (what you were trying to do) may help in some special cases, but it isn't always useful... $\endgroup$ – A.P. Mar 26 '15 at 20:57
  • $\begingroup$ Monomial order is necessary to even define division of two multivariate polynomials, and without division defined it will be difficult to factor stuff. $\endgroup$ – mathreadler Mar 26 '15 at 21:06
  • $\begingroup$ Puiseux series. $\endgroup$ – Nathanson Mar 26 '15 at 21:17
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In this case, it's going to be taking advantage of what you know, to bootstrap yourself into new territory. You basically have

$$y^2 + By + C,$$

where $B = x^2 + 2x - 2$ and $C = x^3 - x^2 - 2x\ $ don't depend on $y$; it looks a little bit like a quadratic in $y$.

Using what we know about quadratics, we "know" it should factor as $(y + p)(y + q) = y^2 + (p + q)y + pq$, if it can be factored. Now, since you know how to factor quadratics with a single variable, that means you know we're looking for a pair of things (it would be a pair of numbers, if we had a good old quadratic with one variable), call them $p$ and $q$, such that $pq = C$ and $p + q = B$.

So, let's see how would could find solutions $p, q$ to $pq = C = x^3 - x^2 - 2x$.

We can factor $$C = x^3 - x^2 - 2x = x(x^2 - x - 2) = x(x - 2)(x + 1).$$

More specifically since we need a pair, let's see what two polynomials multiply to $C = x(x - 2)(x + 1)$: \begin{align*} C &= x(x^2 - x - 2) \\ &=(-x)(-x^2 + x + 2)\\ &= (x^2 - 2x)(x + 1)\\ &= (-x^2 + 2x)(-x - 1)\\ &= (x^2 + x)(x - 2)\\ &= (-x^2 - x)(-x + 2) \end{align*}

So, we have a few possibilities for $p$ and $q$, if we only require $pq = C$. Now let's see if any of them add up to $B = x^2 + 2x - 2$.

Lo and behold, $p = x^2 + x$ and $q = x - 2$ will work. We already know they multiply to $C$, and their sum is indeed $x^2 + 2x - 2 = B$.

Thus, your polynomial factors as

$$(y + p)(y + q) = (y + x^2 + x)(y + x - 2).$$

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Comment on previous answers, with $A=1,$ $B = x^2 + 2x-2,$ $C = x^3 - x^2 -2x$ we get $$ B^2 - 4 AC = x^4 + 4 x^2 + 4 = (x^2 + 2)^2, $$ so, in the quadratic formula, $\sqrt {B^2 - 4AC} = x^2 + 2.$ The "roots" for $y$ are $$ \frac{-B \pm \sqrt {B^2 - 4AC}}{2A}, $$ so the factors are $$ \left( y - \frac{-B - \sqrt {B^2 - 4AC}}{2A} \right) \left( y - \frac{-B + \sqrt {B^2 - 4AC}}{2A} \right), $$
$$ \left( y + \frac{B + \sqrt {B^2 - 4AC}}{2A} \right) \left( y + \frac{B - \sqrt {B^2 - 4AC}}{2A} \right), $$ $$ \left( y + \frac{x^2 + 2 x - 2 + x^2 + 2}{2} \right) \left( y + \frac{x^2 + 2 x - 2 - x^2 - 2}{2} \right), $$ $$ \left( y + \frac{2x^2 + 2 x }{2} \right) \left( y + \frac{ 2 x - 4 }{2} \right), $$ $$ \left( y + x^2 + x \right) \left( y + x-2 \right). $$

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It is not as complicated as many people think. You already noticed that this is a quadratic polynomial in y with coefficients in Z[x]. Vieta's theorem can be applied so we have

$$(y-p_1(x))(y-p_2(x))$$ if it can be factored.

So $p_1(x)$ is a factor of $(x^3−x^2−2x)$. The factors of $x^3−x^2−2x$ are $x$, $x+1$ and $x-2$ multiplied by some constant numbers. You can check the possible solutions.

As soon you have found a factor $(y-p_1(x))$ you can found the remaining factor $p_2(x)$ by division too.

You can use this technic for polynomials with arbitrary number of variable as long all exponents are smaller than or equal to 3.

Here is an example with 3 variables and powers of 3.

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