What skill do I lack to factor multivariate polynomials?

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

Solution 1:

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

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

Solution 2:

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 $$ \left( y - \frac{-B - \sqrt {B^2 - 4AC}}{2A} \right) \left( y - \frac{-B + \sqrt {B^2 - 4AC}}{2A} \right), $$
$$ \left( y + \frac{B + \sqrt {B^2 - 4AC}}{2A} \right) \left( y + \frac{B - \sqrt {B^2 - 4AC}}{2A} \right), $$ $$ \left( y + \frac{x^2 + 2 x - 2 + x^2 + 2}{2} \right) \left( y + \frac{x^2 + 2 x - 2 - x^2 - 2}{2} \right), $$ $$ \left( y + \frac{2x^2 + 2 x }{2} \right) \left( y + \frac{ 2 x - 4 }{2} \right), $$ $$ \left( y + x^2 + x \right) \left( y + x-2 \right). $$

Solution 3:

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.