## College Algebra 7th Edition

$x=0$ $x=-2-\sqrt{2}$ $x=-2+\sqrt{2}$
We factor out $x^2$: $x^{4}+4x^{3}+2x^{2}=0$ $x^{2}(x^{2}+4x+2)=0$ $x^{2}=0$ or $(x^{2}+4x+2)=0$ Left solution: $x=0$ Right solution: use the quadratic formula $x^{2}+4x+2=0$ $x=\displaystyle \frac{-4\pm\sqrt{4^{2}-4*1*2}}{2(1)}=\frac{-4\pm\sqrt{16-8}}{2}=\frac{-4\pm\sqrt{8}}{2}=\frac{-4\pm 2\sqrt{2}}{2}=-2\pm\sqrt{2}$ So the solutions are: $x=0$, $x=-2-\sqrt{2}$ and $x=-2+\sqrt{2}$