Find the real zeros (if any) of the polynomial ?

Jun 23, 2018

$x = 0$ and $x = - 8$

Explanation:

At least $y = 0$ when $x = 0$, since $x = 0$ gives $a {x}^{5} = 0$ for any a.

Other than that you will have to have ${x}^{6} + 9 = 0$, i.e.

${x}^{6} = - 9$
This has a complex solution ($x = {\left(- 9\right)}^{\frac{1}{6}}$ (the 6th root of -9),
which would not be a real zero.

My answer, therefore, would be $x = 0$ and $x = - 8$