Rational Zero Test of Polynomials

Question

Say I'm trying to find the factors of

$r^3 - 3r^2 + 3r - 4 = 0$

I thought I could do the rational zero test by finding the factors of p and q and then doing $\frac{p}{q}$

So, in this case, $p = \pm 4, \pm 2, \pm 1$ and q = $\pm 1$ so the possible factors are $\pm 4, \pm 2, \pm 1$ However, in this example, none of these are factors and the only factor is 2.442.

How would I know this is the case? I thought the rational zero test always works. Is there some condition that must be met that is not met in the example I gave?

Thank you.

Answer 1

The test you are referencing is a way of deciding whether or not there are rational zeros of a polynomial.

You are correct in stating that the only real solution of this equation is $1+3^{1/3}$ (which is approximately 2.442); if there were rational solutions, they would be of the form $\frac{p}{q}$ where $p,q$ are as you described.

Checking all possibilities of $\frac{p}{q}$, namely $4,-4,2,-2,1,-1$, we find none are zeros of the polynomial; thus there are no rational solutions.

I hope this helps.

Answer 2

The Rational Root Theorem or, as you called it, the Rational Zero Test only tells you when the polynomial has a rational root.

As you've tried all the combinations and found that none gives you an actual root, then the polynomial has no rational roots. By the Fundamental Theorem of Algebra however, the polynomial still have 3 roots (counting multiplicities). So, they should be either irrational or complex.

To be precise, the real root of this polynomial is:

\begin{equation} x_1 = 1 + \sqrt[3]{3} \end{equation}

and the other two complex roots are:

\begin{equation} x_2 = 1 - \frac{\sqrt[3]{3}}{2}(1 - i \sqrt{3}), \quad x_3 = 1 - \frac{\sqrt[3]{3}}{2}(1 + i \sqrt{3}). \end{equation}