Explanation:
the non-real zeros (containing i-terms) are created by square roots (of negative numbers). these always come in pairs like the last 2 given zeroes : -5 + i, -5 - i.
(a)
so, we are missing the complementary zero for 2 - 3i :
2 + 3i
(b)
a polynomial of the degree n has n zeros (some can have the same value, when e.g. the curve only touches the x-axis, and is not really intercepting it).
so, in our case, there are 8 zeros.
with the list of given zeros and the result of (a) we have 4 non-real and 1 real zeros.
that means, as we know already 4+1 = 5 zeros of the max. possible 8, the can be only 3 more real zeros max.
so, the max. number of real zeros is 1+3 = 4.
(c)
as mentioned, the non-reals have to come in pairs.
with the list of given zeros and the result of (a) we have 4 non-real and 1 real zeros.
there are 8 zeros in total.
we saw in (b) that we have 5 out of these 8 zeros already. that leaves 3.
because of the pair principle, that gives us only one option für the max. possibilities : 2 more non-real zeros and 1 more real zero.
so, the max. number of non-real zeros is 4+2 = 6.