## Another Princeton Review error

Forum for the GRE subject test in mathematics.
joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

### Another Princeton Review error

At the top of p.15 in the 3rd edition, it states: "[I]rrational roots of rational-coefficient polynomial equations must occur in conjugate radical pairs." However, this fails to account for irrational roots not of the form $s+t\sqrt{u}$.

Following their line of reasoning may prove disastrous. For instance, question #17 of GR0568 asks to find the number of real roots of $p(x)=2x^5+8x-7$. Since $p(0)<0$ and $p(1)>0$, there is at least one root. Moreover, that root must be irrational, as a quick application of the rational roots theorem shows. If we take PR at their word, we're forced to conclude that there are at least two zeroes. However, $p'(x)>0$, so there can be at most one zero. Indeed, the answer key confirms there is exactly one.

Studying with this guide has been nothing but a crap shoot. It seems I'm only slightly more likely to learn useful information than I am to fall prey to subtle yet fatal errors. After two revisions, obvious pitfalls remain. What is going on here?

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Re: Another Princeton Review error

There is no contradiction here. The theorem concerns only roots in form
$a+\sqrt{b}$
while given equation has roots in form
$a + \sqrt{b} + \sqrt[5]{c}$
for which theorem does not apply.

joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

### Re: Another Princeton Review error

Agreed. The problem is not with the theorem, but with Princeton Review's summary of it. We're led to believe that all irrational roots of polynomials take the form
$s+t\sqrt{u}$
and appear in conjugate pairs. Obviously, this is false, as demonstrated here.

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Re: Another Princeton Review error

Agree. The name "irrational roots theorem" sounds a bit misleading tending to make generalization.