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How can we prove that a polynomial only has rational roots when we know the coefficients and the degree? For instance, in illustration, how would we show this for $x^8 +2x^7+3x^6+5x^5+4x^4+3x^3+2x^2+x-1$?

afedder
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1 Answers1

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Hint: By the Rational Root Theorem, the only rationals that could conceivably be roots are $\pm 1$. Plug in. Neither works.

In general, if we have a polynomial $P(x)$ with integer coefficients, where $P(x)=a_0x^n+\cdots +a_n$, where $a_0\ne 0$, $a_n\ne 0$, then the only conceivable rational roots of $P(x)$ are of the form $\dfrac{a}{b}$, where $a$ is a divisor (possibly negative) of $a_n$ and $b$ is a positive divisor of $a_0$.

So there is a finite list of candidates.

If we try them all, and nothing works, there are no rational roots. That's what happened in our concrete case.

To check whether there are any non-rational roots, find all the rational roots. We do have to check for multiple roots, so there is a need for some care.

For example, we can use the Rational Root Theorem to show that the only rational root of $x^3-3x^2+3x-1$ is $1$. However, since our polynomial is $(x-1)^3$, the number $1$ is a triple root of the polynomial.

If the sum of the number of rational roots (counting multiple roots according to their multiplicity) is not $n$, then there are (possibly non-real) roots that are not rational.

Remark: The Rational Root Theorem is more useful in "made up" problems (homework, tests) than in "real life." Often in a problem, when you have say a cubic, and you need the roots, the numbers will have been chosen so that there is a reasonably simple rational root. The same is often true of quadratics in high school. Not so much later, since the Quadratic Formula gives a general procedure for solving quadratic equations.

André Nicolas
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  • First of all, I don't know what the rational roots theorem is. Second of all, how would you prove this generally? I have a few more examples that I need to prove and I don't know how to begin. – afedder Dec 07 '12 at 02:00
  • I have added a link to the Rational Root Theorem. Perhaps after a working for a bit you can add one or even two of the polynomials that are giving you trouble. I will try to look out for it, certainly will if I get a message. – André Nicolas Dec 07 '12 at 02:07
  • Can add a proof of Rational Root Theorem if you are interested in the proof, but probably you can find one, by searching or by hinking about it. If you want to prove it, assume the root has shape $a/b$ where $a$ and $b$ have no factor $\gt 1$ in common, and multiply by $b^n$ to clear denominators. – André Nicolas Dec 07 '12 at 02:11
  • So each rational root ONLY depends on $a_0$ and $a_n$? – afedder Dec 07 '12 at 02:19
  • Knowing $a_0$ and $a_n$ brings you down to a finite, usually small, list of candidates. Nothing else can work, but most, maybe all, the candidates will usually not work either. Whether a candidate works or not is very influenced by the other coefficients $a_i$. In your example, the only rationals that could possibly be roots were $1$ and $-1$. Plugging in, we see neither works, so the polynomial of your post has no rational roots. – André Nicolas Dec 07 '12 at 02:31
  • What I meant is, first and foremost, the only factors that a rational root depends on is those of $a_0$ and $a_n$? In other words, there is no rational roots that might "slip through the cracks" of this possible list? More generally, would there be any exceptions to the application of this theorem? – afedder Dec 07 '12 at 03:11
  • A theorem is a theorem. Nothing can "slip through the cracks". – Robert Israel Dec 07 '12 at 03:14
  • For polynomials with integer coefficients, no exceptions. Of course we need $a_n\ne 0$. If $a_n=0$, like in $2x^4-3x^3+x^2$, we note the rational double root $x=0$, and we are down to $2x^2-3x+1$. – André Nicolas Dec 07 '12 at 03:16
  • Got it, thanks! – afedder Dec 07 '12 at 03:19
  • Is there a way to get the results of this theorem in order to use it without actually proving it in detail from the link? In other words, can you derive these results about possible rational roots by just considering $\mathbb{Z}[x]$ as a UFD? – afedder Dec 07 '12 at 05:34
  • I don't see the point of using more advanced stuff, this is a high-school theorem. Substitute $a/b$ for $x$, multiply by $b^n$ to get integers. Then $a$ divides the first $n$ terms, so it must divide $b^n a_n$. But $a$ and $b$ are relatively prime, so $a$ divides $a_n$. Similarly, $b$ divides the last $n$ terms, so $b$ divides $a_0 a^n$. But $a$ and $b$ are relatively prime, so $b$ divides $a_0$. – André Nicolas Dec 07 '12 at 06:10
  • The reason to use more advanced stuff is because that is the course I am in, modern algebra. The point is that I already understand this theorem under the typical "high school" guise. However, it another context, it can be harder to grasp. – afedder Dec 07 '12 at 10:37