How should I have known that $x^4-2x^3-7x^2+10x+10=(x^2-2x-2)(x^2-5)$?

Question

How should I have known that $$x^4-2x^3-7x^2+10x+10=(x^2-2x-2)(x^2-5)$$?

I was asked to find the splitting field of $f(x)=x^4-2x^3-7x^2+10x+10$. The solution that I was given starts off by noting the given factorization of $f(x)$ into quadratics. Hw should I have seen this factorization? I tried writing $f(x)$ as the product of two arbitrary monic quadratics and matching coefficients but things got messy quite quickly.

In general, to find the splitting field of a quartic polynomial, if all else fails, I believe I could find the roots of the quartic using the general method for solving a quartic by radicals (although I have not learned about this method, or the method for cubic's, I know they exist). Is using the general method for solving quartics a common approach to finding splitting fields of quartics?

Answer 1

The odd part is $-2x(x^2-5)$, so if $x^2-5|x^4-7x^2+10$, or equivalently $y-5|y^2-7y+10$ (which it does), you're lucky enough you can pull out a quadratic factor without a linear term. And having separated the even and odd parts as thus, we know the other factor is $-2x+x^2-2$.

Answer 2

If $f(x):=x^4-2x^3-7x^2+10x+10=p(x)\cdot q(x)$ for monic integer polynomials $p,q\in\mathbb Z[x]$, then we have that $p(0)\cdot q(0)=10$, so we know the constant term of $p$ and $q$ are $\pm1,\pm2,\pm5,$ or $\pm10$.

First observe that $f(\pm1),f(\pm2),f(\pm5),f(\pm10)$ are all nonzero, so neither $p$ nor $q$ has degree $1$. Thus, we conclude that both $p$ and $q$ have degree $2$, say $p(x)=x^2+ax+b$ and $q(x)=x^2+cx+d$ with $a,b,c,d\in\mathbb Z$. We have $a+c=-2$ and $bd=10$.

Moreover, reducing the polynomial modulo $5$, you obtain that $x^4-2x^3-7x^2=\overline p(x)\overline q(x)$. If $\overline p,\overline q\equiv0\pmod x$, then this means $b,c\equiv0\pmod 5$, so that $10=bc\equiv0\pmod{25}$, which is certainly not true. Thus, it must be the case that say, $\overline p(x)=x^2$ and $\overline q(x)=x^2-2x-7=x^2+3x+3\in\mathbb F_5[x]$.

These conditions translate to $a,b\equiv0\pmod 5$ and $c,d\equiv3\pmod 5$. Now you can conclude that $b=-5$ and $d=-2$. Let $a=5k$ and look at the equation

$$x^4-2x^3-7x^2+10x+10=(x^2+5k-5)(x^2+(-2-5k)x-2).$$ Now one may readily solve for $k$.

Answer 3

You know, Im not sure how to answer your question. I too would have been stumped. There are a few things I might have tried.

---

One thing is to depress the quartic by eliminating the cubic term with an $x \rightarrow x + \frac12$ substitution, to arrive at $x^4 - \frac{17}{2}x^2 +2x + \frac{209}{16}$, then employ Rene Descartes' Quadratic Factorization Method for Quartics. This method would give you a set of equations that would help you find your factors. The method typically involves solving a cubic but it may not be necessary to go that far. You may find sufficient information at an earlier stage of the game to complete the task. I have an answer here if youd like to learn the method (Is there a general formula for solving Quartic (Degree $4$) equations?)

---

A second thing I might have tried would be this... Ive never seen this technique written anywhere, I sort of came up with it myself in the early days of school when I was tasked with solving polynomials. Its a trivial substitution when all said and done but Id be darned if anyones ever explicitly suggested this to me. To test for irrational roots of the form $\sqrt{a}$, where $a$ is rational but not a square, simply substitute $x\rightarrow \sqrt{x}$. This method only works for roots of this form.

When simplified you can rearrange your original equation and arrive at $(x^2 - 7x + 10) - 2(x-5)\sqrt{x}$. Notice that since $x$ is rational but $\sqrt{x}$ is irrational, by definition/assumption, we know that $x^2-7x+10$ is rational and that $-2(x-5)\sqrt{x}$ is irrational, having separated the equation into a rational and an irrational component. In order for the entire equation to equal $0$, each component must do independently. And each component is solvable, and is at most a quadratic, having reduced your $n$-degree polynomial to at most an $\lceil n/2 \rceil$-degree polynomial.

The task at hand, now, is to find a solution of $x$ that satisfies both the rational and the irrational components simultaneously, and this is simply the intersection of the sets for each. The irrational component is easy enough to determine: its a linear equation and so $x=5$ is a root. We need only test to verify it is a root of the rational component as well, which it is. Thus $x=5$ is a root to the equation after the $x\rightarrow\sqrt{x}$ substitution, implying that $x=\sqrt{5}$ is a root to your original quartic polynomial.

By way of the fact all coefficients are rational, you are guaranteed to have quadratic conjugate pairs, so $x=-\sqrt{5}$ is a root as well. Thus $x^2-5$ must be one of your quadratic factors.

For the record, this technique can similarly be used for purely imaginary roots of the form $bi$, where $b$ is real.

Using synthetic division you can arrive at your other factor.