How to factor and find zeros of $2x^6+5x^4-x^2$ In the equation $y=2x^6+5x^4-x^2$, how can I factor it in order to find the zeros or x-intercepts? This is what I've gotten to, but I can't see what to do next: $x^2(x^2(2x^2+5)-1)$

$$y=2x^6+5x^4-x^2 = x^2(2x^4 + 5x^2 - 1)$$
For the factor $\;2x^4 + 5x^2 -1,\;$ put $t = x^2$: 
$$2x^4 + 5x^2 - 1 = 2t^2 + 5t - 1$$
Suggestion: Use the quadratic formula to find the roots (also zeros): $t_1, t_2$, then back substitute, knowing $t = x^2 \implies x = \pm \sqrt t$. So for each zero $t_i$, there are two corresponding zeros, $x_j, x_k$ such that $x_{j, k} = \pm t_i$.
