Question

A function has a degree of 8. It has two x-intercepts, one of multiplicity 1 and one of multiplicity 3. How many imaginary zeros does the function have?

Answer 1

two or four

Explanation:

Any degree-X polynomial can have up to 8 zeros (some of them can have multiciplity higher than 1).

The polynomial can be written as:

f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)(x-f)(x-g)(x-h)

and we're already told that there are 2 x-intercepts, one with multiplicity 1. Let's assign those x-intercepts a and b=c=d (multiplicity 3).

We're left with (x-e)(x-f)(x-g)(x-h), where e, f, g and h need to be non-real complex numbers (otherwise they'd provide x-intercepts).

The complex zeros (more accurate term than imaginary) must come in pairs - if a number c is a zero of the function, its conjugate is a zero too. That means that we need to have an even number of complex zeros.

We can either have 4 distinct complex zeros, or 2 complex zeros, each with multiplicity of 2.