This problem came up last week when I was teaching my students about the roots of a polynomial function.

We were talking about the rational root theorem and I gave my class the example
x^3 + x^2 + 4x + 8

We listed all 8 possible rational roots and then checked to see which one was a root. However, after using the Remainder Theorem with each possible root, none of them worked.

Next I went to the calculator and tried to find a zero using that. When I did this, I get 1.607521767.... which appears to me that the zero is irrational. But irrational zeros are supposed to appear in pairs and since this function only crosses the x-axis once, it can't be irrational.

So now I am stuck scratching my head trying to figure out why this doesn't work. I asked several of my coworkers and they do not know why either.

So what am I doing wrong? Is there a theorem that I have forgotten about that tells me what I am doing wrong?

4/27/2009 8:08:35 PM

imaginary roots

4/27/2009 8:30:32 PM

The imaginary roots come in pairs. the real root is its own complex conjugate and thus doesn't need a pair

4/27/2009 8:45:03 PM

Dealing with polynomials from the reals to the reals, nobody ever said that irrational roots come in pairs. For example, the polynomial x^3-2 has just one real root, and it is irrational.

Now, if you take a polynomial with real coefficients, complex roots will always come in pairs: a+bi is a root of a polynomial with real coefficients if and only if a-bi is, too (conjugate pairs). I expect that's what you're thinking of.

4/27/2009 8:47:16 PM

nvmd...

[Edited on April 27, 2009 at 8:50 PM. Reason : nvmd]

4/27/2009 8:48:28 PM

^

ahh,nice edit

4/27/2009 8:51:25 PM

Thankfully my students won't have to find the exact solution to this problem. They won't run into a problem like this on the Algebra 2 EOC.

However, the interesting thing is that the book says that if you are given the roots of a polynomial and one of them is irrational, that you are to assume that another root has to be its conjugate.

4/27/2009 8:58:11 PM

Was the book referring to quadratic polynomials specifically? If you have a quadratic polynomial with rational coefficients then, yes, irrational roots come in conjugate pairs (this should be clear from the quadratic formula). However, this is not true in general for polynomials of higher degree.

4/27/2009 9:59:25 PM

Nope, it mentions all of them. However all of the problems that fall in that category end up being even degree polynomials. There is no mention in the book that if it is an odd degree you can't assume that is true(about the irrational conjugate pairs)

4/28/2009 6:25:49 PM

Just to be clear: Not all even degree polynomials have that property, either. (x^3-2)(x-1) has degree 4, but has exactly one real irrational root (no irrational conjugate to pair it up with).

4/28/2009 11:31:40 PM

you could use p/q theorem

4/29/2009 2:08:32 AM

Sounds like an error in your textbook, which is certainly a possibility.

4/30/2009 11:30:05 PM

in general, just because your calculator returned a root of 1.607521767 doesn't mean it isn't rational (though it certainly may be), it just looks irrational.

your calculator probably uses the secant or chord method to numerically approximate the root, and whatever the solution is it is certainly corrupted with round off error. it may not be a lot, but it is certainly there.

that's right, i blamed round off.

5/20/2009 12:17:37 AM

of course, in this case like others have mentioned, you have a complex pair of roots and a single real.

lookie: http://www25.wolframalpha.com/input/?i=x%5E3+%2B+x%5E2+%2B+4x+%2B+8

5/20/2009 12:19:15 AM

Well, if the polynomial is x^3 + x^2 + 4x + 8 and your calculator gives 1.607521767... as a root, it is actually guaranteed to be irrational. A simple application of the rational root theorem shows that any rational roots must be of the form 1, 2, 4, or 8 (or their corresponding negatives). 1.607521767... obviously isn't one of those, so it's guaranteed to be irrational.

5/20/2009 12:27:54 AM

Yeah, after I went and did more research on it. I realized that I was assuming that irrational roots appear in conjugate pairs.

5/20/2009 7:24:47 PM