Can you find the zeros for
f(x)=2x^4+17x^2+8?

2x⁴+17x²+8

Let X = x²     →→(x = + or -√X )

The quadratic becomes : 2.X² + 17.X + 8. Now solve for X:

X' = [-b+√(b²-4ac)]/2a             and      X" = [-b-√(b²-4ac)]/2a   
Plug I the values of a, b and c and you will find;

X' = -1/2   and X" = - 8
Since X = √x → →X' = √( -1/2) [imaginary roots]
and X" = √-8 →→ X" = 2√-2  [imaginary roots]
Then:

1st ROOT: X' = - √1/2.i  
2nd ROOT: X" = + √1/2.i
3rd ROOT: X" = + 2√2.i
4th ROOT: X" = - 2√2.i