Descartes’ rule of signs:
f(x) has 2 changes of sign, implying at most two positive zeroes.
f(-x)=3x⁴+5x³-x²+8x+4 has 2 changes of sign, implying at most two negative zeroes.
The degree of f(x) is 4, so the number of complex zeroes could be 4-(2+2)=0.
In actual fact, there are 2 complex zeroes and two real positive zeroes.
Complex zeroes always come in conjugate pairs, so in this case there are none, 2 or 4. With two positive zeroes we have 2 negative zeroes or 2 complex zeroes.