I'm having some troubles with it:
$$D=\{(a,b,c,d,e)\in\mathbb{R}^5 \mid \exists x\in\mathbb{R} : ax^4+bx^3+cx^2+dx+e=0\}$$ Prove that ${(1,2,-4,3,-2)}\in \operatorname{int}(D)$
Might use some hints, thanks
$x\in int(D) : if\ there\ is \ a \ r>0\in \mathbb{R} \ that\ B_r(x)\subset D $
$D$ is a subset of $\mathbb R^5$ that each vector from $D$ solves the $4$ degree poly of $x,$ $x$ is just a number in $\mathbb R$ (might means that each vector in $D$ leads to factorization? not sure), I think $\operatorname{int}(D)$ might be just all the vector in $D.$ I'm thinking that they might want me to show that $(1,2,−4,3,−2)$ is extremum of some what by using Lagrange Multipliers, not sure what are my functions $f$ and $F$
So I came up with the solution eventually, hope it is right, it goes like this: First, let us be aware that for $\overline p_0=(1,2,-4,3,-2)\in\mathbb{R^5}\ x=1$ solves the polynomial $p(x)=x^4+2x^3+-4x^2+3x-2$ thus $\overline p_0\in D$ Now lets' define a function $F:\mathbb{R^6}\rightarrow \mathbb{R} \\F(a,b,c,d,e,x)=ax^4+bx^3+cx^2+dx+e$
It is easy to see that $F$ is $C^1(\mathbb{R^6}\times \mathbb{R}$) and $F(\overline p_0,1)=F((1,2,-4,3,-2),1)=0$
$D(F)=(\nabla F(a,b,c,d,e,x))^t=(x^4,x^3,x^2,x,1,4ax^3+3bx^2+2cx+d)^t$ with $rank(D(F))=1$ for all $(a,b,c,d,e,x)\in \mathbb{R^6} $ including our $(\overline p_0,1)$.
now let us apply the implicit function theorem and it gives us that there exist a domain $V \times W \subset \mathbb{R^5} \times\mathbb{R}$ of $(\overline p_0,1)$ and a function $g\in C^1(V,\mathbb{R})\ ,\ g(V)\subset W $ in which that:
$g(\overline p_0)=1$
$\forall (\overline p,x)\in(V\times W) \mid F(\overline p,x)=0 \Leftrightarrow g(\overline p)=x \Leftrightarrow F(\overline p,g(\overline p))=0$
and the second claim implies us that we've found a domain $V$ of $ \ \overline p_0$ in $D$.