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.
