x5+3x4-3x3-9x2-10x-30 looks like it can be factorised:
x4(x+3)-3x2(x+3)-10(x+3)=(x4-3x2-10)(x+3)=
(x2-5)(x2+2)(x+3)=
(x-√5)(x+√5)(x-i√2)(x+i√2)(x+3), from which all zeroes (real and complex) can be seen.
A degree 5 polynomial must have at least one real zero (because 5 is an odd number and complex roots come in pairs). The possibilities for the remaining zeroes are:
4 real zeroes;
2 real zeroes, and 2 complex zeroes (conjugates);
4 complex zeroes, (2 pairs of conjugate zeroes).
Rational zeroes come from the factors of 30 = (1, 2, 3, 5). However, having factorised the polynomial, the zeroes are:
√5, -√5, i√2, -i√2, -3. So there are 3 real zeroes and 2 complex zeroes. Amongst the real zeroes there is a pair of irrational zeroes. Only one rational zero (-3) appears in the list of actual zeroes. The other zeroes contain an irrational number (√2 or √5, and 2 and 5 are two of the rational zeroes).