Some polynomials with real coefficients, like $x^2 + 1$, have no real zeros. As it turns out, every polynomial with a complex coefficient has a complex zero. Every polynomial of odd degree with real coefficients has a real zero.

The Fundamental Theorem

The fundamental theorem of algebra says that every non-constant polynomial in a single variable $z$, so any polynomial of the form

$c_nx^n + c_{n-1}x^{n-1} + \ldots c_0$

where $n > 0$ and $c_n \not = 0$, has at least one complex root. 

There are lots of proofs of the fundamental theorem of algebra. However, despite its name, no purely algebraic proof exists, since every proof makes use of the fact that $\mathbb{C}$ is complete. 

In particular, since every real number is also a complex number, every polynomial with real coefficients does admit a complex root. For example, the polynomial $x^2 + 1$ has $i$ as a root.

Alternative Statement

Saying that $x_0$ is a root of a polynomial $f(x)$ is the same as saying that $(x-x_0)$ divides $f(x).$ We say that a root $x_0$ has multiplicity $m$ if $(x-x_0)^m$ divides $f(x)$ but $(x-x_0)^{m+1}$ does not. For example, the polynomial 

$x^4(x-i)^3(x+\pi)$

admits one complex root of multiplicity $4$, namely $x_0 = 0$, one complex root of multiplicity $3$, namely $x_1 = i$, and one complex root of multiplicity $1$, namely $x_2 = - \pi$. The sum of the multiplicity of the roots equals the degree of the polynomial, $8$. For non-zero complex polynomials, this turns out to be true in general and follows directly from the fundamental theorem of algebra. 

Indeed, a polynomial of degree $0$ takes on the form $c_0$, where $c_0 \not = 0$, and thus has no zeros.

For a general polynomial $f(x)$ of degree $n$, the fundamental theorem of algebra says that we can find one root $x_0$ of $f(x)$. Thus we can factor $f(x)$ as

$f(x) = (x-x_0)f_1(x)$

where $f_1(x)$ is a non-zero polynomial of degree $n-1.$ So if the multiplicities of the roots of $f_1(x)$ add to $n-1$, the multiplicity of the roots of $f$ add to $n$. 

So since the property is true for all polynomials of degree $0$, it is also true for all polynomials of degree $1$. And since it is true for all polynomials of degree $1$, it is also true for all polynomials of degree $2$. In general, for any $n \in \mathbb{N}$, we will be able to conclude that the property is true for all polynomials of degree $n.$ Thus the property is true for all polynomials.

Conversely, if the multiplicities of the roots of a polynomial add to its degree, and if its degree is at least $1$ (i.e. it is not constant), then it follows that it has at least one zero.

So an alternative statement of the fundamental theorem of algebra is:

The multiplicities of the complex roots of a nonzero polynomial with complex coefficients add to the degree of said polynomial.

The Complex Conjugate Root Theorem

The complex conjugate root theorem says that if a complex number $a+bi$ is a zero of a polynomial with real coefficients, then its complex conjugate $a-bi$ is also a zero of this polynomial.

Now suppose our real polynomial admits a root $a+bi$ with $b \not = 0$. By dividing with the real polynomial$(x-(a+bi))(x-(a-bi))=(x-a)^2 +b^2$, we obtain another real polynomial, for which the complex conjugate root theorem again applies. In this way, we see that the total multiplicity of non-real complex roots of a polynomial with real coefficients must always be even. 

This last remark, together with the alternative statement of the fundamental theorem of algebra, tells us that the parity of the real roots (counted with multiplicity) of a polynomial with real coefficients must be the same as the parity of the degree of said polynomial. Therefore, a polynomial of even degree admits an even number of real roots, and a polynomial of odd degree admits an odd number of real roots (counted with multiplicity). In particular, every polynomial of odd degree with real coefficients admits at least one real root.$$