Demonstrate that this equation has three real roots if fulfills this condition I have this: demonstrate that this equation 
$$ {x}^{3}+p\,x+q=0 $$
has one real root if "p" is positive and has three real roots if 
$$ \frac{{q}^{2}}{4}+\frac{{p}^{3}}{27}<0 $$
I did demonstrate that it has one real root, but the three real roots I don't know how to manage to demonstrate them and I don't have idea how to approach it.
I'm in first year calculus.
 A: Here is a solution using concepts from analysis. Let $f(x)=x^3+p\,x+q$ and $p<0$. If the equation has three roots, then $f$ has a local maximum and a local minimum, and the local minimum has to be negative. The derivative of $f$ has two roots:$\pm\sqrt{-p/3\,}$. The negative one corresponds to the maximum and the positive one to the minimum. The condition for the existence of three roots is then $f(\sqrt{-p/3\,})<0$.
Edit: you also want the local maximum to be positive, that is, $f(-\sqrt{-p/3\,})>0$. Thanks go to Arthur for pointing this out.
A: This comes quite directly, using Cardano's famous method for cubics. Starting with your equation, let $$\begin{align}u+v &= x \\ 3uv + p &= 0\end{align}$$
Now, substituting in the original cubic (and simplifying): $$ u^3 + v^3 +(3uv + p)(u+v) + q = 0$$
We see that $$\begin{align}u^3 + v^3 &= -q \\ u^3v^3 &= \frac{-p^3}{27} \end{align}$$
From this we know that $U = u^3,\, V = v^3$ satisfy the degree two equation $$X^2+qX-\frac{p^3}{27} = 0$$
Studying the discriminant of this equation and undoing the change of variables will get you the result you wish.
A: The simpler is to use symmetric functions of the roots $x_0,x_1$ of the derivative: by the intermediate value theorem, $f(x)$ has $3$ real roots if and only if $f(x_0)f(x_1)<0$.
It's easy to compute $f(x_0)f(x_1)$ as a symmetric function of $x_0$ and $x_1$:
$$f(x_0)f(x_1)=(x_0x_1)^3+px_0x_1(x_0^2+x_1^2)+q(x_0^3+x_1^3)+p^2x_0x_1+pq(x_0+x_1)+q^2.$$
Now, $x_0$ and $x_1$ are the two (opposite) roots of $-\dfrac p3$, so that
$$x_0x_1=\frac p 3, \quad x_0+x_1=0,\quad x_0^2+x_1^2=-\frac{2p}3. $$
Replacing in $f(x_0)f(x_1)$, you'll find the condition:
$$\frac{p^3}{27} -\frac{2p^3}9+\frac{p^3}3+q^2=\frac{4p^3}{27}+q^2<0,$$
which you can rewrite as$$\Bigl(\frac p3\Bigr)^3+\Bigl(\frac q2\Bigr)^2<0.$$
A: Let's $\alpha$ is any real root of $f(x)=x^3+px+q$,then 
$$
f(x)=(x-\alpha)(x^2+ax+b).
$$
From this we get $\alpha=a$, so $$
f(x)=(x-a)(x^2+ax+b),
$$
and 
$$
b-a^2=p,-ab=q.
$$
Now we put these equations in 
$$
\frac{{q}^{2}}{4}+\frac{{p}^{3}}{27}<0
$$
and get 
$$
4a^6-12ba^4-15a^2b^2-4b^3=(b+2a^2)^2(a^2-4b)\gt0,
$$
so $a^2-4b\gt0$ and our polynomial has three real roots.
