Let $f(x)$ be any quartic polynomial with coefficients from $\{ -1, +1 \}$. Replacing $f(x)$ by $-f(x)$ if necessary, we can assume $f(x)$ is monic. i.e.
$$f(x) = x^4 + ax^3 + bx^2 + cx + d\quad\text{ with }\quad a,b,c,d \in \{ -1, +1 \}$$
If $f(x)$ has $4$ real roots $\lambda_1,\lambda_2,\lambda_3,\lambda_4$, then by Vieta's formula, we have
$$\sum_{i=1}^4 \lambda_i = -a, \sum_{1\le i < j\le 4} \lambda_i\lambda_j = b
\quad\text{ and }\quad\prod_{i=1}^4 \lambda_i = d$$
Notice
$$\sum_{i=1}^4 \lambda_i^2 = \left(\sum_{i=1}^4\lambda_i\right)^2 - 2\sum_{1\le i < j \le 4}\lambda_i\lambda_j = a^2 - 2b = 1 -2b$$
Since $\sum_{i=1}^4 \lambda_i^2 \ge 0$, we need $b = -1$. As a result, $$\sum_{i=1}^4 \lambda_i^2 = 3$$
By AM $\ge$ GM, this leads to
$$\frac34 = \frac14\sum_{i=1}^4 \lambda_i^2 \ge \left(\prod_{i=1}^4 \lambda_i^2\right)^{1/4} = (d^2)^{1/4} = 1$$
This is impossible and hence $f(x)$ cannot has 4 real roots.
Let $f(z)$ be a degree $3$ complex polynomial with real coefficients.
... tells you precisely that you can think of $f$ as a function $f\colon\mathbb R\to\mathbb R$ and can use intermediate value theorem.
After you obtain one real root $\alpha$, divide $f$ by $x-\alpha$ to get a degree $2$ polynomial. I assume you know how to prove how many roots it has.
Best Answer
Assume that $P$ has degree $n$ and let $x_1,x_2,\dots,x _n$ be all its roots (repetitions are allowed). Then $P(x)=c\prod_{k=1}^n (x-x_k)$, and if $x$ is not a root of $P$ we have that $$\frac{P'(x)}{P(x)}=\sum_{k=1}^n \frac{1}{x-x_k}.$$ After taking the derivative we obtain $$\frac{P''(x)P(x)-(P'(x))^2}{(P(x))^2}=-\sum_{k=1}^n \frac{1}{(x-x_k)^2}.$$ Finally by letting $x=a$ (which is not a root) we get a contradiction: $$0=\frac{P''(a)P(a)-(P'(a))^2}{(P(a))^2}=-\sum_{k=1}^n \frac{1}{(a-x_k)^2}<0$$ where the right-hand side is negative because $a, x_1,x_2,\dots,x _n$ are all real.