The possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are [tex]\mathbf{\pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}[/tex]
How to determine the possible zeros?
The function is given as:
f(x) = 3x^6 + 4x^3 -2x^2 + 4
The leading coefficient of the function is:
p = 3
The constant term is
q = 4
Take the factors of the above terms
p = 1 and 3
q = 1, 2 and 4
The possible zeros are then calculated as:
[tex]\mathbf{Zeros = \pm\frac{Factors\ of\ q}{Factors\ of\ p}}[/tex]
So, we have:
[tex]\mathbf{Zeros = \pm\frac{1,2,4}{1,3}}[/tex]
Expand
[tex]\mathbf{Zeros = \pm\frac{1,2,4}{1},\pm\frac{1,2,4}{3}}[/tex]
Solve
[tex]\mathbf{Zeros = \pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}[/tex]
Hence, the possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are [tex]\mathbf{\pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}[/tex]
Read more about rational root theorem at:
https://brainly.com/question/9353378
#SPJ1
