Answer:
[tex]x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}[/tex]
Step-by-step explanation:
We are given the function:
[tex]f(x)=-4x^5-8x^3+12x[/tex]
And we want to finds its zeros.
Therefore:
[tex]0=-4x^5-8x^3+12x[/tex]
Firstly, we can divide everything by -4:
[tex]0=x^5+2x^3-3x[/tex]
Factor out an x:
[tex]0=x(x^4+2x^2-3)[/tex]
This is in quadratic form. For simplicity, we can let:
[tex]u=x^2[/tex]
Then by substitution:
[tex]0=x(u^2+2u-3)[/tex]
Factor:
[tex]0=x(u+3)(u-1)[/tex]
Substitute back:
[tex]0=x(x^2+3)(x^2-1)[/tex]
By the Zero Product Property:
[tex]x=0\text{ and } x^2+3=0\text{ and } x^2-1=0[/tex]
Solving for each case:
[tex]x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}[/tex]
Therefore, our real and complex zeros are:
[tex]x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}[/tex]
