Answer
Makes sense.
Work Step by Step
The Rational Zero Theorem states that
if $\displaystyle \frac{p}{q}$ (where $p$ is reduced to lowest terms) is a rational zero of $f$,
then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
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1. So, if there are rational zeros $\displaystyle \frac{p}{q}$, then
$p;\quad \pm 1, \pm 2$
$q:\quad \pm 1$
possible $\displaystyle \frac{p}{q}:\quad \pm 1, \pm 2.$
There are four, so the first part of the statement is true.
2. The second part can be checked in several ways:
Evaluate f(possible root) ... see if any yield 0,
test the possible $\displaystyle \frac{p}{q}$ with synthetic division,
graph and analyze - see if there are any intersections with the x-axis,
or, like we do here, algebraically conclude that, for real x:
1. $x^{4}$ and $x^{2}$ are never negative (even exponents),
2. $x^{4}+3x^{2}+2$ is never less than 2, so
3. f(x) is never zero ... f has no real roots.
Verdict: makes sense.
