How many real zeros does y = 3x 4 - 2x 2 + x - 3 have?

A. 5
B. 4
C. 3
D. 2
my answer is d

y = 3x 4 - 2x 2 + x - 3

factor

(3x^4 -2x^2) (x-3)

Factor out x^2 from the first, so you have

(x^2) (3x^2 - 2) (x-3), so

x^2=0

3x^2 - 2 = 0

x-3=0

So how many zeros can you get after you solve these?