Is it true that a 3rd order polynomial must have at least one real root? While solving a problem a friend said - this polynomial is $3^{rd}$ order ($ax^3+bx^2+cx+d$), with $\{a,b,c,d\}$ real coefficients, so it must have a real root. I didn't want to sound stupid and I said sure.
I can't figure out if he's right. Is he right? Can someone help me with this?
Edit: Earlier I had asked if it must have a negative root. It's the real root that he said such polynomial must have.

If $p(x) = ax^3 + bx^2 + cx + d $ and if $a$ and $d$ happen to be of same sign, then you can conclude that there must be at least one negative root. Probably this is what your friend would have done as well.
EDIT
The question was changed from having a negative root to there is at least one real root.
This is a consequence of the fact that $p(x)$ is continuous and the intermediate value theorem for continuous functions.
The important observation is that $$\lim_{x \rightarrow \infty} p(x) = \text{sign(a)} \cdot \infty$$
and
$$\lim_{x \rightarrow -\infty} p(x) = -\text{sign(a)} \cdot \infty$$
For instance, if $a > 0$, then $\displaystyle \lim_{x \rightarrow \infty} p(x) = \infty$
and $\displaystyle \lim_{x \rightarrow -\infty} p(x) = -\infty$.
Since the function is continuous, by intermediate value theorem, it must hit all values between $- \infty$ and $\infty$ for some $x$. Hence, in particular, it must hit $0$. Intutively, the function is positive for some large $x$ and negative for some large $x$ and since the function has no jumps, it must hit $0$ somewhere on the real axis before it changes sign.
