Linear algebra| finding a base

You want to find a basis of the polynomials of degree $4$ for which $p(-1)=0$ and you want the $x$-coefficient of each polynomial in the basis to be $200$. There is one more condition in your question, namely $p^n(-1)=0$, but this one doesn't make sense (what is $n$?).

We will consider $4$ polynomials:
$$p_1(x) = x^4+bx^3+cx^2+200 x+ e$$
$$p_2(x) = x^3 + c' x^2 + 200x + e'$$
$$p_3(x) = x^2 + 200x + e''$$
$$p_4(x) = 200 x + e'''$$

For any values of the unknown coefficients these polynomials are linearly independent. Thats relatively easy to see as in the standard basis or $\Bbb R_4[x]$ given by $1, x, x^2, x^3,x^4$ these polynomials have the representation
$$\begin{pmatrix}e\\ 200\\ c\\ b\\ 1\end{pmatrix}, \begin{pmatrix}e'\\ 200\\ c'\\ 1\\0 \\ \end{pmatrix}, \begin{pmatrix}e''\\ 200\\ 1\\0\\0\end{pmatrix},\begin{pmatrix}e'''\\ 200 \\ 0\\ 0\\0\end{pmatrix}$$
So any linear combination $\lambda_1 p_1 +\ldots \lambda_4p_4$ of these will only have $0$ in the $x^4$ component if the $\lambda_1=0$, then the $x^3$ component is just $\lambda_2$, which can only be $0$ if $\lambda_2=0$ and so on, so you get
$$\lambda_1p_1+\ldots \lambda_4p_4 =0 \iff \lambda_1=\lambda_2=\lambda_3=\lambda_4=0$$
and they are linearly independent.

It remains to find appropriate coefficients:
$$p_1(-1) = 1-b+c-200+e\overset!=0$$
so taking $b=c=0$ and $e=199$ gives an element in $U$.
$$p_2(-1) = -1+c'-200+e'\overset!=0$$
so taking $c'=0$ and $e'=201$ gives an element in $U$.
$$p_3(-1)=1-200+e''\overset!=$$
so taking $e''=199$ gives an elment in $U$.
$$p_4=-200+e'''\overset!=0$$
so taking $e'''=200$ gives an element in $U$.
In total the polynomials are then:

$\qquad p_1(x) = x^4+200x+199$
$\qquad p_2(x) = x^3 +200x +201$
$\qquad p_3(x)= x^2+200x+199$
$\qquad p_4(x) = 200x+200$

Again you should not forget that there is an additional condition on $U$ that you have not transcribed correctly, thats why this answer gets a basis of $4$ (rahter than $3$) elements.