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I've given my class an example: $$2x^3+3x^2+6x+4=0$$ By the rational root theorem, if there is a rational root then it should be of the form $\frac{p}{q}$ where $p$ is a factor of 4 and $q$ is a factor of 2.
The problem I'm having is this, none of the possible candidates $\{\pm\frac1{2}\pm1,\pm2,\pm4\}$ are rational roots. I'm thinking because $\frac{p}{q}$ is not in simplest form is the reason (as it is stated in the RRT, $\frac{p}{q}$ must be in simplest form...) So what am I to do to find the rational root? There's clearly a rational root here since it is a cubic right?

Eleven-Eleven
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  • check: http://www.wolframalpha.com/input/?i=solve+2+x%5E3%2B3+x%5E2%2B6+x%2B4%3D0 – janmarqz Jan 02 '14 at 18:56
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    Actually, there is no obligatory there will be a rational root. I guess the most simple example would be $x^3+2=0$... – Galc127 Jan 02 '14 at 18:56
  • There are plenty of polynomial equations which have no rational root. Take $x^2+1=0$ fro example. Now if a polynomial has odd degree then it must have a real root by the Intermediate value theorem, but that real root need not be rational. – Wintermute Jan 02 '14 at 18:58
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    The theorem says that if there is a rational root, then it must be of some prescribed form. But it does not say that there must be a rational root. http://en.wikipedia.org/wiki/Rational_root_theorem – fkraiem Jan 02 '14 at 19:00
  • That's good, @Galc127. I was thinking along those lines too. So now, how does one solve the above cubic? Do I have to start using the Cardano transformation and go from there? – Eleven-Eleven Jan 02 '14 at 19:00
  • That was my thoughts after rereading the theorem, @fkraiem. It also stated that p and q must be relatively prime as well, but it's been a while and I didn't want to give my class the wrong information. – Eleven-Eleven Jan 02 '14 at 19:01
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    @ChristopherErnst So what you have done is a proof that there are no rational roots. If there were one, it would be of the prescribed form, but none of the numbers of that form are roots. – fkraiem Jan 02 '14 at 19:04
  • @ChristopherErnst, if you want to find the exact solutions of the cubic equation there is no avoidance of using Tartaglia-Cardano formula. If approximation is good enough there are a lot of linear approximations (such Newton-Repson, etc...) – Galc127 Jan 02 '14 at 19:04
  • Did you just make this example up on your own? If so, and you didn’t get it from a source, then the chances are that it has no roots and is irreducible. You can exclude all positive possibilities, as well as $-1$, because that sum is odd. You can exclude $-4$ because that sum is congruent to $4$ modulo $8$. Looks as if the other possibilities are no good either. – Lubin Jan 02 '14 at 19:04
  • @Lubin, yeah. My original equation was actually $x^3$, not $2x^3$, but I told my students if that were the case, then another possible root was $\pm\frac{1}{2}$. Then I tried solving and it was impossible. Then I went back and reread the theorem and chalked it up to $\gcd(2,4)\neq{1}$. Now I'm just looking for methods to solving – Eleven-Eleven Jan 02 '14 at 19:10
  • Well, you may have heard of the Cubic Formula. But NEVER teach this to high-school students. – Lubin Jan 02 '14 at 19:13
  • Yeah, I've seen the cubic formula...i was hoping for a simpler method for high schoolers, but that's probably why it's not in the text!! Cheers everyone! – Eleven-Eleven Jan 02 '14 at 19:14
  • The really simple answer is that only rational roots show up using that theorem. Real numbers that aren't rational will not necessarily show up. – Zediiiii Aug 10 '18 at 00:32

1 Answers1

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The rational root theorem constrains all rational roots of a polynomial.

For your equation:

$$2x^3+3x^2+6x+4=0$$

all rational roots of this equation must be of the form $p/q$ (in lowest terms) where $p$ divides $4$ evenly, and $q$ divides $2$ evenly.

Your possible candidates are indeed $\{\pm\frac1{2}\pm1,\pm2,\pm4\}$. The only real root of this equation, however, is $\frac{1}{2}(-1 + \sqrt[3]{3} - \sqrt[3]{9})$. This root is obviously irrational since neither $3$ nor $9$ are perfect cubes.

Since there are no rational roots of this equation, there are no roots to be constrained by the rational root theorem.

More strongly (and more correctly), because none of the candidate values satisfy the equation, there are no rational roots.

The existence of candidate roots from the rational root theorem does not mean that there are any rational values that satisfy the equation. It only says that, if there are indeed rational roots that satisfy the equation, they must be taken from the list of candidates.

John
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  • You wrote: "... is $\frac{1}{2}(-1 + \sqrt[3]{3} - \sqrt[3]{9})$. This root is obviously irrational." How is this obviously irrational? The simplest way I can think of to prove this is to prove the cubic version of the rational root theorem and then apply it to $\sqrt[3]{3} - \sqrt[3]{9};$ [cube both sides of $x = \sqrt[3]{3} - \sqrt[3]{9} = \sqrt[3]{3}\left(1 - \sqrt[3]{3}\right)$ to get $x^3 = 3(1 - 3x - 3),$ or $x^3 + 9x + 6 = 0,$ and then apply rational root test to this equation], but your context suggests there is a simpler way. – Dave L. Renfro Jan 02 '14 at 20:04
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    @DaveL.Renfro It is obviously irrational and the proof here is simple. It is obvious that it is real and one can check it is a root of the polynomial. Then one only need consider if it is rational or irrational. If it were rational, it would have appeared on the list provided by the Rational Roots Theorem. Since it is clearly not, then this number is irrational. – mathematics2x2life Jan 02 '14 at 20:06
  • @mathematics2x2life: As I stated, your context suggests there is a simpler way (but this may have gotten lost in the bracketed comment just before it). What is somewhat obvious (and perhaps intended, but the wording doesn't suggest this to me) is that the expression is not equal to any of the rational root candidates, but even for this I think some not entirely trivial numerical estimates need to be proved to make this air-tight. – Dave L. Renfro Jan 02 '14 at 20:13
  • @DaveL.Renfro I think you are making a mountain out of a mole hill. Given the level of question, I would believe that it would not even be required to show that $\frac{1}{2}(-1+\sqrt[3]{3}-\sqrt[3]{9})$ is irrational. However, showing that it is not any of the rational roots listed by the Rational Roots Theorem is also a trivial matter if one is forced to do it. Moreover, it should be obvious by inspection alone. – mathematics2x2life Jan 02 '14 at 20:28
  • @DaveL.Renfro But again, realize that the answer isn't saying that one need show any of this. The giving of the root was just a bonus. The core of the answer is that none of the choices listed by the Rational Roots Theorem are actually roots of the given polynomial. Therefore, there are no rational roots. If you don't believe that, simply check each of those values and see that they do not make the polynomial zero. – mathematics2x2life Jan 02 '14 at 20:31
  • @mathematics2x2life: My concern is that the way John's comment was worded will lead a nontrivial percentage of high school teachers (and a high percentage of high school students) to think the expression represents an irrational number simply because (1) it involves radicals, and (2) there seems to be no way to rewrite it as a quotient of integers. I agree that The core of the answer is the none of the choices listed by the Rational Roots Theorem are actually roots. – Dave L. Renfro Jan 02 '14 at 20:55
  • @mathematics2x2life: By the way, I also agree that my comment ... is that the expression is not equal to any of the rational root candidates, but even for this I think some not entirely trivial numerical estimates need to be proved to make this air-tight was neither here nor there (one of the down-sides of trying to quickly slip in comments/posts where I work). – Dave L. Renfro Jan 02 '14 at 20:59
  • Tried to clarify my answer to reduce confusion. – John Jan 02 '14 at 21:53
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    The cube root of 3 has degree 3 over the rationals, so no rational linear combination of it, its square, and 1, can be zero. Thus, $(1/2)(-1+\root3\of3-\root3\of9$ is "obviously" irrational. – Gerry Myerson Jan 02 '14 at 21:55
  • @Gerry Myerson: Nice! (Because I happen to know the degree $3$ fact can be proved in a fairly straightforward high school algebraic way.) – Dave L. Renfro Jan 02 '14 at 22:47