If $F= {\mathbb{Q}[x]}/{(x^5+5x^2-10)}$ then show that $[F:\mathbb{Q}]=5$.
Now $f(x)=x^5+5x^2-10$ is $5$-Eisenstein, so irreducible in $\mathbb{Z}[x]$ so by gauss lemma, also irreducible in $\mathbb{Q}[x]$. Now what will be the intermediate fields? Any help is appreciated
By the Tower Law . See Dummit and Foote page 523 . If $L$ such that $\Bbb{Q}\subseteq L\subseteq F$ be a subfield of $F$. Then $[F:\Bbb{Q}]=[F:L]\cdot [L:\Bbb{Q}]$ . Now as $5$ is a prime number , we have either $[F:L]=5\implies [L:\Bbb{Q}]=1\implies L=\Bbb{Q}$ or $[F:L]=1\implies L=F$ .
In all the cases above we are using the fact that if a subspace $W$ of a vector space $V$ has dimension equal , i.e if $\dim(W)=\dim(V)$ then $V=W$.
Proof that $[F:\Bbb{Q}]=5$.
Consider the polynomial $f(x)=x^{5}+5x^{2}-10\in \Bbb{Q}[x]$ and the quotient map $q:\Bbb{Q}[x]\to \mathbb{Q}[x]/(x^5+5x^2-10)$ .
Then $q(x)=x\pmod{f(x)}$ and call this as $\bar{x}$. Now as $\Bbb{Q}[x]$ is an Euclidean Domain we have for any polynomial $g(x)\in\Bbb{Q}[x]$ , $g(x)=a(x)f(x)+r(x)$ for some $a(x),r(x)$ such that $\deg(r(x))<5$ .
It then follows that any element in $\mathbb{Q}[x]/(x^5+5x^2-10)$ is of the form $r(x)\pmod {f(x)}$ with $\deg(r(x))<5$ as the quotient map is surjective.
This means that in the quotient we can represent any element by $\displaystyle\sum_{i=0}^{4}a_{i}\bar{x}^{i}$ . Thus the set $\{1,\bar{x},...,\bar{x}^{4}\}$ is a spanning set for the vector space $\mathbb{Q}[x]/(x^5+5x^2-10)$ over $\Bbb{Q}$.
Now it only remains to prove that this set is linearly independent.
If possible let there exists $c_{0},...,c_{4}$ not all $0$ such that $c_{0}+c_{1}\bar{x}+...+c_{4}\bar{x}^{4}=0$ . This means that $f(x)$ must divide $c_{0}+c_{1}x+...+c_{4}x^{4}=0$ but this is impossible as $f(x)$ has degree $5$ and the above is of degree $4$. This means that the only possibility was that $c_{0}=c_{1}=...=c_{4}=0$.
Thus $\{1,\bar{x},...,\bar{x}^{4}\}$ is a linearly independent spanning set and hence a basis and thus $F$ as a vector space over $\Bbb{Q}$ has dimension $5$.