Let $$A = \begin{pmatrix} 0 & & & -x_0 \\ 1 & 0 & & -x_1\\ & 1 & 0& ... \\ & &... & -x_{n-1} \end{pmatrix} $$
determine the characteristic polynom and decide wether A is diagonalizable. I know that the formula for determining c.p is $det(A-\lambda E)$ but how can I use this for the given matrix?
As dxiv pointed out, the matrix is the companion matrix of the polynomial $$p(y)=y^n+x_{n-1}y^{n-1}+\cdots+x_1y+x_0$$
which is the characteristic polynomial of the matrix. Mostly, the exercise goes in the other direction : We have a polynomial and want to find a matrix having this characteristic polynomial. One matrix is the so-called companion-matrix.
Hint:
Prove $$\begin{vmatrix} -\lambda&0&0&\dots&-x_0\\ 1&-\lambda&0&\dots&-x_1\\ 0 & 1 &-\lambda&\dots&-x_2\\ \vdots&\vdots&\vdots&&\vdots\\ 0&0&0&\vdots&--x_{n-1} \end{vmatrix}=(-1)^n(\lambda^n+x_{n-1}\lambda^{n-1}+\dots+x_1\lambda+x_0)$$ by induction on $n$, expanding this determinant along the first row.