Answer
$x=0$
$x=4-2\sqrt{3}$
$x=4+2\sqrt{3}$
Work Step by Step
We factor out $y^3$:
$y^{5}-8y^{4}+4y^{3}=0$
$y^{3}(y^{2}-8y+4)=0$
$y^{3}=0$ or $(y^{2}-8y+4)=0$
Left solution: $y=0$
Right solution: use quadratic formula ($a=1, b=-8, c=4$):
$y^{2}-8y+4=0$
$y=\displaystyle \frac{-(-8)\pm\sqrt{(-8)^{2}-4*1*4}}{2(1)}=\frac{8\pm\sqrt{48}}{2}=4\pm 2\sqrt{3}$
So the solutions are: $x=0$, $x=4-2\sqrt{3}$ and $x=4+2\sqrt{3}$
