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how to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x​

Sagot :

XKelvinidn

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]

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