The answer is 5(a-2)(a^2+5)
Let's call P(a)=5a^3−10a^2+25a−50
First, you notice that 5 divides all the coefficients of P(a), so P(a)=5Q(a) with Q(a)=a^3-2a^2+25a-50
This is factoring by grouping, you collect a 5 outside the polynomial because you notice it divides every coefficient.
Then, if you want a prime factorisation, you can notice that 2 is a root of Q(a)
So (a-2) divides the polynomial, so you have, using Ruffini's rule
P(a)=5Q(a)=5(a-2)(a^2+5) and you know it's a prime factorisation because (a-2) has degree 1 and (a^2+5) doesn't have any roots in QQ (nor in RR).
If you want the factorisation in CC, you have that (a^2+5)=(a+5i)(a-5i), but I think this is a little out of your league.
PS: you notice 2 is a root just trying, e.g. in the beginning I thought Q(a) was irreducible, but I wasn't able to prove that with any criterion, and I found out that 2 is a root when I was looking where the polynomial was positive or negative, hoping Gauss' lemma could prove it was irreducible in ZZ so in QQ.
There's Cardano's formula for roots of polynomials of degree 3, but it's horrible and I usually avoid it.