If a is algebraic of degree 5 over then by definition there is a minimal irreducible polynomial of degree n such that a is a root ie p(a)=0.

But then consider the polynomial q(x)=p(x+5).

q(a-5)=p((a-5)+5)=p(a)=0.

It clearly also has degree n, and by the hint it seems you have a theorem to show you why q(x)=p(x+5) is also irreducible.

If you cannot find it, a proof by contradiction is very easy, suppose q(x) were reducible, then you would have a factorization of p(x) which would make p(x) not irreducible which is a contradiction. Thus you have a minimal irreducible polynomial of degree n for which a-5 is a root, thus a-5 is algebraic of degree n.