Let $D$ be the set of all $3-tuples$, $(a,b,c)$ in $R^3$ such that the cubic polynomial $x^3+ax^2+bx+c$ has three real roots, i.e., $D=\{(a,b,c) \in \mathbb{R}^3 \mid x^3+ax^2+bx+c \textit{ factors into linear factors over } \mathbb{R} \}$. Then show that $D$ is a connected subset of $\mathbb{R}^3$ with the standard topology.
I have no clue whatsoever for this problem. Can someone suggest me how to approach this problem? I found some NASC conditions for a cubic to have three real roots but I think there might be some elegant way to solve this.
If we can show that $D$ is path connected then we are through. Another approach might be expressing this set as a continuous image of some known connected set. I am unable to proceed using only these ideas.
If $\alpha<\beta<\gamma$ then the coefficients $a,b,c$ of $(x-\alpha)(x-\beta)(x-\gamma)$ depend continuously on $\alpha,\beta,\gamma$. But the triple $(\alpha,\beta,\gamma)$ can be deformed to $(-1,0,1)$ continuously while preserving the order in the triple of distinct roots. Therefore the set is connected.