Apply the definition of a discriminant Problem Statement
Denote a polynomial $p(x)$ of degree $n>1$ with $n$ real roots. Prove that all roots of $p'(x)$ are real. 
Progress
The definition of the discriminant of a polynomial of degree $n$ is given by 
$a_n^{2n-2}\prod_{i<j}{(r_i-r_j)^2}=(-1)^{n(n-1)/2}a_n^{2n-2}\prod_{i \neq j}{(r_i-r_j)}$ where $a_n$ is the leading coefficient and $r_1,r_2...,r_n$ are the roots. My approach is to use that for $n=k$ the discriminant is non-negative and use that to prove that it implies that $n=k-1$ also is non-negative. However I do not understand this definition since it requires the knowledge of roots beforehand. 
For simplicity: How can I derive the discriminant to second/third grade polynomials using the definition above. Let's say $p_2(x)=a_1x^2+a_2x+a_3$ and $p_3(x)=a_1x^3+a_2x^2+a_3x+a_4$. 
Maybe easier?
Assume a polynomial of degree $n$ and use the factor theorem. $a_1x^{n}+a_2x^{n-1}+a_3x^{n-2}+...+a_{n-1}x+a_{n} = a(x-x_0)(x-x_1)\cdot ...\cdot (x-x_n)$
$P'(x)$ has degree $n-1$ requiring that we divide $P(x)$ by one factor $(x-x_i), i>1$ leading to a polynomial of real roots. Is this argument correct?
 A: Hint: If I give you a polynomial $p$ of degree $n$ with $n$ real roots, where are the roots of $p'$ going to be? Think along the lines of the mean value theorem, or even the special case of Rolle's theorem.
A: This is a special case of the Gauss–Lucas theorem. You have a nice proof in this Wikipedia article.
Edit
Since you wanted a comment on your attempts, here it is.
For your discriminant approach, there is not much to say. I don't see how can you prove what you want, as the roots change, and you'd need to relate the discriminant of your polynomial with the discriminant of its derivative.
The second attempt is more interesting and was going in the direction of the first thing I thought of. However, it is wrong. You claim:

Assume a polynomial of degree $n$ and use the factor theorem. $a_1x^{n}+a_2x^{n-1}+a_3x^{n-2}+...+a_{n-1}x+a_{n} = a(x-x_0)(x-x_1)\cdot ...\cdot (x-x_n)$
$P'(x)$ has degree $n-1$ requiring that we divide $P(x)$ by one factor $(x-x_i), i>1$ leading to a polynomial of real roots. Is this argument correct?

This is wrong, because when deriving a product, we derive one element and leave the others, and then sum these bits. So,
$$P'(x) = a \sum_{i=1}^n \prod_{\substack{j=1 \\ i \ne j}}^n (x-x_j) = \sum_{i=1}^n \frac{P(x)}{x-x_i}.$$
Since we have a sum, I see no way to relate the roots of $P'(x)$ with $P(x)$ using this approach.
If some of this is not clear enough, feel free to ask in the comments.
A: Normally when a hypothesis says there are $n$ things of some kind, one may interpret this as saying there are $n$ distinct such things; after all, we have all have learned that in counting, one shouldn't visit the same object more than once lest one get a wrong answer. Interpreting the question this way the answer is easy: ordering the $n$ roots by size, the $n-1$ pairs of adjacent roots define as many bounded intervals, the interiors of which are disjoint, and by Rolle's theorem each interior contains a point where the derivative $p'(x)$ vanishes; these $n-1$ (real) roots of $p'$ are all it can have, given that $\deg p'=n-1$.
But counting roots of a polynomial$~p$ is a bit special, and often when poeple counts roots, a root$~r$ is implicitly counted with its multiplicity as root of$~p$ (the number of times $p$ can be successively divided by $x-r$). With this interpretation the hypothesis it does not exclude that $p$ has multiple roots (it just says that all complex roots of $p$ are in fact real), so there can be less than $n$ distinct roots, and the above proof does not work. It is not hard to repair though. Set $m$ to the sum over all roots of one less than the multiplicity of the root. Then $p$ has are $n-m$ distinct roots, and the above argument gives us (only) $n-m-1$ distinct roots of$~p'$. But the contribution of a root$~r$ to$~m$, if nonzero, is precisely the multiplicity of $r$ as a root of the derivative$~p'$, so counting these with multiplicity gives is $m$ more roots of$~p$, all distinct from the ones we found first because those all lie in the interior of an interval between roots of$~p$. That gives overall $(n-m-1)+m=n-1$ (real) roots of$~p'$ counted with multiplicity, and again that is all that is can have.
A: As rghthndsd said we can use rolle's theory: Let $p(x)\in\mathbb R[x]$a polynom whose degree is n and has n real roots. If $p$ has $n$ real roots, it can be written as $p(x)=\displaystyle\prod_{i=1}^n(x-r_i)$ where $r_1...r_n\in \mathbb R$ are its roots. Rolle's theorm gives us that $\forall i\neq j $ (without loss of generality $j>i$) exist $t_a\in\mathbb R$  ($1\le a\le n-1$) s.t $p\prime(a_t)=0$. That also means that $p\prime(x)$has $n-1$ real roots. If $p\prime(x)$ has at least another root, say complex $a+b\cdot i$, then $deg(p\prime(x))$ would be at least n (for $n-1$ roots from original polynom we add at least one complex root), in contradiction to the fact that $deg(p\prime(x))=n-1$.
The main point is that between ech two roots of polynom exist a maximum/minimum point (which its x-value is roots of the derivative).
EXAMPLE: Assume we take the polynom $p(x)=x^5 - 26\cdot x^4 + 246x^3 - 1036x^2 + 1865x - 1050$. Let's plot it: 
As one can see it has 5 roots (in $x=1,3,5,7,10\in\mathbb R$). Now we can simply derive it to get $p\prime(x)=5x^4 - 104x^3 + 738x^2 - 2072x + 1865$.Ploting the derivative, we can notice it has 4 real roots in $x=1.75,4,6.25,9$ all in $\mathbb R$.  as required.
 
