Let p be a prime and a be an element of Zp. Show that x^p+a is reducible in Zp[x]

im not sure if this is the best way to approach the question but

as x^p+a is reducible there must be a factor

therefore x^p+a=0

and x^p=-a for some x

would it then suffice to say that as Zp contains a primitive root,

hence there exists an element of Zp congruent to -a, and therefore there is a root in Zp? This doesn't seem like a very good solid solution