Answer
An odd degree polynomial MUST have a real zero.
An even degree polynomial CAN have no real zeros.
Work Step by Step
see Properties of Roots of Polynomial Equations (page 382)
1. If a polynomial equation is of degree $n$, then counting multiple roots separately, the equation has $n$ roots.
2. If $a+bi$ is a root of a polynomial equation with real coefficients $(b\neq 0)$, then the imaginary number $a-bi$ is also a root.
Imaginary roots, if they exist, occur in conjugate pairs.
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For odd degree polynomials, the roots can not all be complex, since complex roots come in conjugate pairs.
If there is one complex root, then there is another, its conjugate pair.
The leftover, third root is not paired, hence it is not complex. It must be real.
Conclusion: select any odd degree polynomial. There WILL be (at least) 1 real zero.
On the other hand, if the degree is even, n=2k, we COULD have k pairs of complex conjugate zeros... which leaves no real zeros.
Examples:
$f(x)=x^{2}-1$ has two zeros,$ (\pm 1)$
$f(x)=x^{4}-1$ has two real zeros, $\pm 1$,
$($so it has another two which are not,
that is, two which are complex ($\pm i$)
$f(x)=x^{2}+1,$
$f(x)=x^{4}+1$
$f(x)=x^{6}+1$ etc
have no real zeros (all come in complex conjugate pairs).
