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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?

ACB
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  • Thank you, fixed –  Jan 14 '22 at 16:56
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    Mathematica: Factor[x^4 - 2 x^3 - 7 x^2 + 10 x + 10]. $$\left(x^2-5\right) \left(x^2-2 x-2\right)$$ – David G. Stork Jan 14 '22 at 16:57
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    For the sake of intrinsic mathematical understanding, are there pencil and paper ways of figuring this out? –  Jan 14 '22 at 16:59
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    Since these polynomials are for illustrative purposes, you were probably meant to assume that all of the roots are integers, so you could look at the factorizations of $10$ to guess what the roots might be. – Chickenmancer Jan 14 '22 at 16:59
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    I think integers, or some friendly nth roots, roots of unity, or a mix of those. – Chickenmancer Jan 14 '22 at 17:00
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    As for your question... since this was asked in an academic setting, one might hope that it factors "nicely" with integer coefficients. My first instinct would have been to check if $0,1,-1,2,-2$ were roots or if some other root jumps out at me. When that fails, I would likely then assume that it factoring nicely would be done with quadratics and would begin looking at factorizations of the constant term and leading term. As $10=2\times 5=1\times 10=(-2)\times (-5)=\dots$ I would have checked if $(x^2-2),(x^2-5),(x^2+2),(x^2+5)$ is a divisor, hoping for a zero coefficient on $x$. – JMoravitz Jan 14 '22 at 17:00
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    In practice, such nice factorizations don't commonly occur outside of academia, so going straight to a computer program would have been better. – JMoravitz Jan 14 '22 at 17:02
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    https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/auckly29.pdf – John Douma Jan 14 '22 at 17:02
  • What about using the general method for solving quartics to find the roots of $f(x)$? I mean, the quadratic formula is used all the time for quadratics, is the quartic method used for solving quartics in practice? (I understand the quartic method is surely much more tedious...) –  Jan 14 '22 at 17:04
  • @JohnDouma Thanks!! –  Jan 14 '22 at 17:05
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    The quartic formula in its direct form is certainly not usable by hand – Sidharth Ghoshal Jan 14 '22 at 17:13
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    @mathematicalsamurai Maybe they expected $,x^4-2x^3\underbrace{-2x^2-5x^2}_{=,-7x^2}+10x+10,$. – dxiv Jan 14 '22 at 17:14
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    If this is asked in class with the expectation of an immediate answer, it's unfair. If it's posed as a homework exercise, it's certainly OK to ask a CAS. It seems that the purpose here is to find a splitting field once the roots are known (i.e. Galois theory), not to find the roots themselves (i.e. Precalculus). So there is an Precalculus obstacle here that prevents an Algebra student from showing their knowledge. If there was no hint given, I would call this a badly written problem. – Hans Engler Jan 14 '22 at 17:18
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    Experience and guessing. Cross fingers and hope we can factor it as $(x^2 + ax + b)(x^2 + cx + d)=x^4 + (a+c)x^3 + (b + ac +d)x^2 + (bc+ad)x + bd = x^4-2x^3 -7x^2 + 10x + 10$ so $bd=10; (bc+ad)=10; b+ac+d=-7;a+c=-2$. Four equations four unknowns. Regretably not linear equations so there's no guarantee theres solutions but worth trying. We can gamble and assume $10$ factors as $1\cdot 10$ or $2\cdot 5$ and as the coefficients look relatively close we can try assuming $b=\pm 2;d=\pm 5$ and that $2c+5a=\pm 10; ac=-14,0;a+c=-2$. $c=0; a=-2;$ and $b,d$ being negative works nicely. – fleablood Jan 14 '22 at 17:24
  • Apropos of the comments of JMoravitz and dxiv, perhaps a bit of "lore" to keep tucked away is that if the monic quartic is a product of two quadratic factors for which one has no linear term, we have the product $$ (x^2 + ax + b) \ · \ (x^2 + c) \ \ = \ \ x^4 \ + \ a·x^3 \ + \ (b + c)·x^2 \ + \ ac·x \ + \ bc \ \ . $$ So here we can "read off" $ \ a \ = \ -2 \ \ , \ \ b + c \ = \ -7 \ \ , \ \ ac \ = \ 10 \ \ , \ $ and, of course, $ \ bc \ = \ 10 \ \ . \ $ We then have more than enough information to find and check that $ \ b \ = \ -2 \ $ and $ \ c \ = \ -5 \ \ . $ –  Aug 02 '22 at 07:33

3 Answers3

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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 $k$.

Kenta S
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  • Did you mean to write $\overline p,\overline q\equiv0\pmod x$?? To me it looks like that should be mod 5 –  Jan 14 '22 at 17:18
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    I meant $\overline p,\overline q\equiv0\pmod x$. Note that we have $x^2|\overline p\cdot\overline q$, we must either have that $x|\overline p$ and $x|\overline q$, or otherwise $x^2|\overline p$ (or $x^2|\overline q$). – Kenta S Jan 14 '22 at 17:20
  • "Thus, we conclude that both p and q have degree 2" If it were me, I'd emphasize assuming the polynomial is factorable. However as you did say "if" in your very first line it could be a moot point. – fleablood Jan 14 '22 at 17:53
  • @fleablood I agree with you, I certainly should've put more emphasis on the "If"; it's a big one. – Kenta S Jan 14 '22 at 18:15
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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 $-2x+x^2-2$.

J.G.
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    What is this voodoo about odd parts and even parts of a polynomial, and that the even part of the odd part divides the even part...? O_o –  Jan 14 '22 at 17:21
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    I think this mean $-2x^3 + 10x$ (the odd powers of $x$) is equal to $-2x(x^2 - 5)$ so $x^4-2x^3 -7x^2 + 10x + 10 = \underbrace{(x^4-7x^2+10)}{\text{"even" part}}+ \underbrace{(-2x^3+10x)}{\text{"odd" part}} +=\frac {x^4-7x^2+10}{x^2-5}(x^2 -5) + -2x(x^2 -5) = (x^2-5)(\frac {(x^2-5)(x^2-2) }{x^2 -5} -2x)=(x^2-5)(x^2 -2 -2x)$. It does require going out on a wing and a prayer that the polynomial is factorable by some $x^2 + d$. – fleablood Jan 14 '22 at 17:37
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    @fleablood It does, yes; but with pedagogical exercises, one is supposed to go out on wings & prayers sometimes. – J.G. Jan 14 '22 at 17:38
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    @mathematicalsamurai Even and odd parts refer to terms of even and odd degree; it's standard terminology. I never said "the even part of the odd part divides the even part"; I said the odd part has an obvious factor, which turns out to divide the even part. – J.G. Jan 14 '22 at 17:39
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    @J.G at least easy ones in which you can abort quickly and not lose much time if the whole thing goes #### up. – fleablood Jan 14 '22 at 17:40
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    It's a standard technique and I really should use and trust it more. It's easy to do. It frequently won't work but it doesn't waste much time and more often than not when it does fail it will give you insight to t slight offset that will work. – fleablood Jan 14 '22 at 17:42
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    @mathematicalsamurai The systematic way to solve this is via undetermined coefficients, as in fleablood's other comment, which is not too terrible in this case. Before you fallback to that, it (only) makes sense to try shortcuts which don't take much time to sort out. These include the odd/even split shown here, or eyeballing the whole thing for a lucky factor per my comment under the main question. – dxiv Jan 14 '22 at 17:47
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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.

SquishyRhode
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