Please! Can some show how Ax^3 + Bx^2 + CX + D has atleast one real root using the Intermediate Value Theorem?
I don't know about Intermediate Value Theorem, but by the fundamental theorem of polynomials, there are as many complex roots as the degree of the polynomial.
So every cubic has 3 complex roots.
But since roots that have an imaginary part always appear as complex conjugates, that means that in this case, at most there are 2 roots with imaginary part. That means there must be at least one real root.
I highly suspect that the proof of your fundamental theorem of polynomials *uses* the fact that any odd degree polynomial has a root in R. At least the only proof I know to prove the fundamental theorem of algebra, which uses a combination of Galois theory and Sylow theory, indeed uses this fact. I would appreciate it if you could have a check and tell me whether it uses this fact or not.
And there's a tiny bit problem in the statement. A should be nonzero otherwise the statement is false.
If A is nonzero, then WLOG a>0. The goal is to find two points x_1,x_2 such that f(x_1)>0 but f(x_2)<0. By IVT, there exists some zero in between. x_1 is chosen as big as possible, and x_2 is chosen as small (negative and the abs is as big) as possible. You can set any bound you like, as long as you can prove f(x_1)>0,f(x_2)<0.
A quicker way to state this (it really is just masking what FancyMouse said, it's quicker if you're allowed to use it) is that $\displaystyle \lim_{x\to\pm\infty}P(x)=\pm\text{sgn } A\text{ }\infty$. And thus, it is clear that you may choose $\displaystyle |T|$ large enough so that (WLOG the other case is analgous) $\displaystyle P(x)<0, x<T$ and $\displaystyle P(x)>0,\text{ }x>T$. Then, apply the IVT.