Prove that $ax^3+bx^2+c = 0$ have less then three real roots on $[0,1]$ I need to know whether what I wrote below makes sense. Plus if so what can be improved.
Suppose the opposite there exist $$g(x) =ax^3+bx^2+c$$ that has 3 real roots in $[0,1]$ and it is not zero polynomial.
If $a = 0$ then $deg(g)\leq2$ so it cannot have three roots.
So $a \neq 0$.
$$g(x) = ax^3 + bx^2 + c$$
There exist $\alpha_1$, $\alpha_2$, $\alpha_3$ $\in [0,1]$ such that $g(\alpha_1)=g(\alpha_2)=g(\alpha_3)=0$.
Polynomial is continuous and differentiable so by Rolle theorem there exist
$\beta_1 \in (\alpha_1, \alpha_2)$, $\beta_2 \in (\alpha_2, \alpha_3)$
$$f'(\beta_1) = 0, f'(\beta_2) = 0$$
$$f'(x) = 3ax^2+2bx$$
$$f'(\beta_i) = 0 = 3a\beta_i^2 + 2b\beta_i$$
$\beta_i = 0$ is a root. For $\beta_i \neq 0$ there is:
$$0 = 3a\beta_i^2 + 2b\beta_i$$
$$-2b = 3a\beta_i + 2b$$
$$\frac{-2b}{3a} = \beta_i$$
$\beta_1 = 0$, $\beta_2 = \frac{-2b}{3a}$. 
but $\beta_1 \in  (\alpha_1, \alpha_2) \subset (0,1)$
Thanks in advance for your time.

A different approach can be as follows. We will argue by contradiction. Assume that there are three different roots in $[0,1],$ say $\alpha, \beta,\gamma.$ Then
$$ax^3+bx^2+c=a(x-\alpha)(x-\beta)(x-\gamma)=a[x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x+\alpha\beta\gamma].$$
So, it must be $$\alpha\beta+\alpha\gamma+\beta\gamma=0.$$ Since  $\alpha, \beta,\gamma\ge 0$ we get that at least two of the three roots must be zero. That is, the polynomyal has at most two different roots, what gives us the desired contradiction.
