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?
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
1 answer