# How do you find all zeros of f(x)=x^5+x^3-6x?

Jan 12, 2017

The zeros of $f \left(x\right)$ are: $0 , \pm \sqrt{2} , \pm \sqrt{3} i$

#### Explanation:

$f \left(x\right) = {x}^{5} + {x}^{3} - 6 x$

The zeros of $f \left(x\right)$ are the values of $x$ where $f \left(x\right) = 0$

That is where: ${x}^{5} + {x}^{3} - 6 x = 0$

$x \left({x}^{4} + {x}^{2} - 6\right) = 0$

Hence $x = 0$ or ${x}^{4} + {x}^{2} - 6 = 0$

Let $z = {x}^{2}$

$\therefore {z}^{2} + z - 6 = 0$

$\left(z + 3\right) \left(z - 2\right) = 0$

$\to z = 2 \mathmr{and} - 3$

$\therefore x = \pm \sqrt{2} \mathmr{and} \pm \sqrt{3} i$

The zeros of $f \left(x\right)$ are: $0 , \pm \sqrt{2} , \pm \sqrt{3} i$