If $f(T)g(T)=0$ and $V_1=g(T)(V)$, $V_2=f(T)(V)$ proof that $V=V_1\oplus V_2$

$\Bbb K$ is a field, $V$ and let $T:V\longrightarrow V$ be a linear map. $\Bbb K[x]$ is the vector space of the polynomials over $\Bbb K$.

Suppose that

$f,g\in \Bbb K[x]$, $f(T)g(T)=0$, $V_1=g(T)(V)$  and $V_2=f(T)(V) \space$; then prove that $$V=V_1\oplus V_2$$ and that $V_1$ and $V_2$are $T-$invariant.

This is what i got so far:

if $$u\in V_1\cap V_2 \Rightarrow u=f(T)(v_1)=g(T)(v_2)$$ $$f(T)(u)=f(T)f(T)(v_1)=f(T)g(T)(v_2)=0$$

I'm trying to proof that the intersection is equal to {0}, and than i would proof $\forall v \in V ;v=v_1+v_2$ with $v_1\in V_1 \wedge v_2\in V_2$

but i wasn't able to do more than that, so if you have any idea, i would be glad to hear it.

This is not true as stated. For example, if $f=g$ then $V_1=V_2$, in which case we certainly cannot have $V=V_1\oplus V_2$ (unless $V=\{0\}$, in which case everything equals $0$...)
It is true if $f$ and $g$ have no common factor. In that case, since the polynomials over a field form a PID there exist polynomials $p$ and $q$ with $$f(t)p(t)+g(t)q(t)=1.$$So $f(T)p(T)+g(T)q(T)=I$, hence for every $x\in V$ we have $$x=f(T)(p(T)x)=g(T)(q(T)x)\in V_1+V_2.$$Now suppose $x\in V_1\cap V_2$, so $x=p(T)v=g(T)w$. Then $$x=f(T)p(T)g(T)v+g(T)q(T)f(T)w=f(T)g(T)(p(T)v+q(T)w)=0.$$
(Note here that if $r$ and $s$ are polynomials it's clear that $r(T)s(T)=s(T)r(T)$.)