Answer:
![x=\{0, -1, 1, -i√(3), i√(3)\}](https://img.qammunity.org/2022/formulas/mathematics/high-school/wkl1unw1a1lbyo6vq2yaw9iczm6wc5gpdx.png)
Explanation:
We are given the function:
![f(x)=-4x^5-8x^3+12x](https://img.qammunity.org/2022/formulas/mathematics/high-school/xlkt8v0xhjg2ze2g5uq1wc337xh7whb80r.png)
And we want to finds its zeros.
Therefore:
![0=-4x^5-8x^3+12x](https://img.qammunity.org/2022/formulas/mathematics/high-school/3lbiy9ngcpa4opeoxw4atc700j0ubq5pne.png)
Firstly, we can divide everything by -4:
![0=x^5+2x^3-3x](https://img.qammunity.org/2022/formulas/mathematics/high-school/8kj992lxbtrtsfmx15p8j3csekkomzrbao.png)
Factor out an x:
![0=x(x^4+2x^2-3)](https://img.qammunity.org/2022/formulas/mathematics/high-school/9ic4175y1pv10qqq9pj07ox4s96fa4wkf5.png)
This is in quadratic form. For simplicity, we can let:
![u=x^2](https://img.qammunity.org/2022/formulas/mathematics/high-school/y69zdcp8jlnlk53bgk3560ceq3h3ryih5k.png)
Then by substitution:
![0=x(u^2+2u-3)](https://img.qammunity.org/2022/formulas/mathematics/high-school/p3hfr10nwf39wxie643i53p9t8qjyfqeta.png)
Factor:
![0=x(u+3)(u-1)](https://img.qammunity.org/2022/formulas/mathematics/high-school/j5kqvufux1iruzeterm1xj68v4b5qiklet.png)
Substitute back:
![0=x(x^2+3)(x^2-1)](https://img.qammunity.org/2022/formulas/mathematics/high-school/5vo9223iqilblkkfvxptnbhb08xiz6jg6s.png)
By the Zero Product Property:
![x=0\text{ and } x^2+3=0\text{ and } x^2-1=0](https://img.qammunity.org/2022/formulas/mathematics/high-school/i1rtnph8t4bsngkfs9vg11jn5st87jpqjo.png)
Solving for each case:
![x=0\text{ and } x=\pm√(-3)\text{ and } x=\pm√(1)](https://img.qammunity.org/2022/formulas/mathematics/high-school/3qymu0lw33hduxd6n86yvt4go802lxh2qv.png)
Therefore, our real and complex zeros are:
![x=\{0, -1, 1, -i√(3), i√(3)\}](https://img.qammunity.org/2022/formulas/mathematics/high-school/wkl1unw1a1lbyo6vq2yaw9iczm6wc5gpdx.png)