Question

Find possible zeroes f(x)=3x^6+4x^3-2x^2+4

Answer 1

The possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are $\mathbf{\pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}$

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:

$\mathbf{Zeros = \pm\frac{Factors\ of\ q}{Factors\ of\ p}}$

So, we have:

$\mathbf{Zeros = \pm\frac{1,2,4}{1,3}}$

Expand

$\mathbf{Zeros = \pm\frac{1,2,4}{1},\pm\frac{1,2,4}{3}}$

Solve

$\mathbf{Zeros = \pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}$

Hence, the possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are $\mathbf{\pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}$

Read more about rational root theorem at:

brainly.com/question/9353378

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