Let $U$ is the set of all polynomials $u$ on field $\mathbb F$ such that $u(3)=u(-2)=0$. Determine the set $W$ such that $P(x)=U\oplus W$. Let $U$ is the set of all polynomials $u$ on field $\mathbb F$ such that $u(3)=u(-2)=0$. Check if $U$ is a subspace of the set of all polynomials $P(x)$ on $\mathbb F$ and if it is, determine a set $W$ such that $P(x)=U\oplus W$.
Attempt:
$U=\{u(x):u(x)\mod (x-3)=0 \land u(x)\mod (x+2)= 0\}$
$U$ is a subspace of $P(x)$ iff
$1)$ $\forall u_1,u_2\in U\Rightarrow u_1+u_2\in U$
$2)$ $\forall u\in U,\forall \alpha\in \mathbb F\Rightarrow \alpha u\in U$
Question: How to check if $U$ is a subspace of $P(x)$?
Assuming $U$ is a subspace of $P(x)\Rightarrow$
$P(x)=U\oplus W=U+W \mod n$ 
where $n$ should be the total number of polynomials in $U$ and $W$.
This means that $W$ is the set of all polynomials $u(x)$ defined as 
$$W=\{u(x): u(3)\neq u(-2)\neq 0 \lor u(3)\neq u(-2)=0\lor u(3)=u(-2)\neq 0\}$$
Is this correct?
 A: First, lets review your attempt:
You state that $U=\{u(x):u(x)\mod (x-3)=0 \land u(x)\mod (x+2)= 0\}$ which, as I understands it, means that $u(x)$ can be written as $a(x-3)$ and as $b(x+2)$ for some $a,b \in \mathbb F$. However, it is NOT equivalent to $u(3)=u(-2)=0$. Therefore, that statement is false, so for now we will just use $U=\{u : u(3)=0  \land u(-2)=0\}$.
Your definition of a subspace is correct, now lets prove that $U$ is a subspace of $\mathbb F[X]$ ($\mathbb F[X]$ is the common notation for the set of all polynomials on $\mathbb F$):
1) $u_1,u_2 \in \mathbb F \iff u_1(3)=0 \land u_1(-2)=0 \land u_2(3)=0 \land u_2(-2)=0 \implies (u_1+u_2)(3)=0 \land (u_1+u_2)(-2)=0 \iff u_1+u_2\in \mathbb F$
2)$u \in \mathbb F \iff u(3)=0 \land u(-2)=0 \implies \alpha u(3)=0 \land \alpha u(-2)=0 \iff \alpha u \in \mathbb F$
Therefore, $U$ is a subspace of $\mathbb F[X]$
Now we want to find some subspace W so that $\mathbb F[X]=U\oplus W$. That means that $U \cap W=\{0\}$ and $\forall p \in \mathbb F[X], \exists u \in U,w \in W : u+w=p$
For that kind of question, there is a general method, but it is rather cumbersome, so we will guess the answer: the only condition we have on $u$ is that $u(3)=0 \land u(-2)=0$, so we want $w$ to have the same values as $p$ at $3$ and $-2$, then we define $u$ as $p-w$. Since $u(3)=u(-2)=0, u\in U$.
Now, how do we construct $W$ so that $U \cap W=\{0\}$ and $\forall a,b \in \mathbb F, \exists w \in W : w(3)=a \land w(-2)=b$ ?
We just have to take $W=Span(\frac{(x+2)}{5},\frac{(3-x)}{5})$, so $w=a\frac{(x+2)}{5}+b\frac{(3-x)}{5}$ fits the conditions and $U \cap W=\{0\}$, since $w\in U \implies a=b=0$. However, keep in mind that this $W$ isn't the only possible answer; is it just one of the (infinitely many) answers.
Feel free to ask for clarifications if you have trouble understanding something.
