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
[Math] What skill do I lack to factor multivariate polynomials
algebra-precalculuspolynomials
Best Answer
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
$$(y + p)(y + q) = (y + x^2 + x)(y + x - 2).$$