prooving that certain point is interior using the implict function theorem

Question

I have a question that I got for homework that I have a difficult time to solve(I need to use the implicit function theorem) $$let \ D=\{(a,b,c,d,e)\in R^5 | \ ax^4+bx^3+cx^2+dx+e=0 \ has \ a \ real \ solution \} $$ $$prove \ that \ (1,2,-4,3,-2) \ is \ an \ interior \ point \ in \ D. \ $$ I've tried to mark Function $F(a,b,c,d,e,x)= ax^4+bx^3+cx^2+dx+e$ and I successfully managed to show it fullfills the implicit function's conditions( $F \in C^1(D \times R,R)$, its differential according to $x$ valued at $(1,2,-4,3,-2,1)$ is of maximum rank, and $F(1,2,-4,3,-2,1)=0)$ ,but what do I do next ? thanks in regards

Answer 1

The theorem then states that there are open sets $U\subset\mathbb{R}^5$, $V\subset\mathbb{R}$, with $(1,2,-4,3,2)\in U$ and $1\in V$, and a function $\varphi:U\to V$ such that $$\forall((a,b,c,d,e),x)\in U\times V,F(a,b,c,d,e,x)=0\iff x=\varphi(a,b,c,d,e).$$ Thus, for each $(a,b,c,d,e)\in U$, the real $x=\varphi(a,b,c,d,e)$ is a root of the polynomial $aX^4+bX^3+cX^2+dX+e$. Then, the set $U$ is an open set included in $D$ containing $(1,2,-4,3,2)$, so this latter is an interior point of $D$.