Question

State how many imaginary and real zeros the function has.
f(x) = x^5 + 7x^4 + 2x^3 + 14x^2 + x + 7

Answer 1

5, because the largest exponent tells you how many zeros there are.

Answer 2

real or imginary

Answer 3

You need to use http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs . This can only estimate, to find definitively, you'll need to find factors.

Using this theory, the sequence of pairs of successive signs is ++, ++, ++,++,++ and thus there are no positive real roots.

To find the number of negative roots, change the signs of the coefficients of the terms with odd exponents f(-x); the sequence of pairs of successive signs is -+,+-,-+,+-,-+. There are 5 sign changes, so there 5, 3 or 1 negative real roots.

Thus, the maximum number of positive real roots is 0, the maximum number of negative real roots is 5 and so the minimum number of complex roots is 5 - (0+5)=0.

In fact, using http://en.wikipedia.org/wiki/Rational_root_theorem we find that one root is x=-7. Using synthetic division, we find that the other factor is x^4+2 x^2+1, which factors as (x-i)^2 (x+i)^2, with roots x=-i and x=i, each of multiplicity 2.

So there are in fact 1 real zero and 2 imaginary zeros, each of multiplicity 2.