Let $f(x)$ be a cubic real polynomial with a leading coefficient of $1$.
Let $\alpha$ be all the complex roots of $f(x)$. $$V=\{y\,|\,y=g(\alpha),g(x) \text{ is a real polynomial}\}$$
Prove that $V$ is a linear space over the real numbers, and find $\dim V$.
I don't know how to connect $f(\alpha)$ and $g(\alpha)$ or $V$, so I'm confused here.
Given any complex root $\alpha$ of $f(x)$, $V$ is a $\mathbb R$-subspace of $\mathbb C$ because $0=f(\alpha)$ so $0\in V$ and if $y_i$ in $V$ for $i=1,2$ there exists $g_i$ real polynomials such that $y_i=g_i(\alpha)$ hence for any $\lambda\in\mathbb R$, $y_1+\lambda y_2=(g_1+\lambda g_2)(\alpha)$ so as $g_1+\lambda g_2$ is a real polynomial we get $y_1+\lambda y_2\in V$. If we denote $F$ the $\mathbb R$ subspace of $\mathbb C$ spanned by $1$, $\alpha$ and $\alpha^2$ we have that $\alpha^3\in F$ thanks to $f(\alpha)=0$ and for any $n\geqslant 3$, $\alpha^n\in F$. So for any real polynomial $g$ we have $g(\alpha)\in F$ thus $V\subset F$. The reverse inclusion holds true as any $y\in F$ can be written $y=a+b\alpha+c\alpha^2$ with $a,b,c$ real numbers that is $y=g(\alpha)$ where $g(x)=a+bx+cx^2$. It follows that $V=F$, if $\alpha\in\mathbb R$ then $V=F=\mathbb R$ and if $\alpha\in\mathbb C\setminus\mathbb R$ then $V=F=\mathbb C$ because $\text{dim}_{\mathbb R}(V)\geqslant 2$ and $\text{dim}_{\mathbb R}\mathbb C=2$ (which you can see because $(1,i)$ is a basis for $\mathbb C$).