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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.
It is true that a cubic polynomial must have a real root. Since the lead coefficient is not $0$, we have that $$ \lim_{x\to-\infty}ax^3+bx^2+cx+d=\left\{\begin{array}{}-\infty&\text{if }a>0\\+\infty&\text{if }a<0\end{array}\right. $$ and $$ \lim_{x\to+\infty}ax^3+bx^2+cx+d=\left\{\begin{array}{}+\infty&\text{if }a>0\\-\infty&\text{if }a<0\end{array}\right. $$ Since a polynomial is continuous, by the Intermediate Value Theorem, if it takes a positive value and a negative value, it must take every value in between, in particular $0$.
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.
Any polynomial $p(x)$ of odd degree satisfies $$ \lim_{x \rightarrow \infty} p(x) = \pm \infty $$ and $$ \lim_{x \rightarrow -\infty} p(x) = \mp \infty. $$ That is, one side of the graph goes up forever and the other side of the graph goes down forever. Since polynomials are also continuous, the graph has no choice but to cross the $x$ axis somewhere, giving a real root. (The fact that the continuous function has "no choice" is formalized by the Intermediate Value Theorem).
No, cubic polynomials must have a real root. But it does not have to be negative. Consider $x^3-1$. Its only real root is the positive number $1$.
Other answers used the intermediate value theorem. Here's an alternative. By the complex conjugate root theorem, non-real roots occur in complex conjugate pairs. By the fundamental theorem of algebra, a cubic polynomial has three complex roots. It follows that a real cubic polynomial must have at least one real root.
