Question

How do you find the zeroes for `f(x)=2x^4-2x^2-40`?

Answer 1

Simplify by replacing `x^2` with `y`;
solve as a standard quadratic for expressions in terms of `y` that result in zeros;
replace `y` in those expressions with `x^2` and solve for values of `x`

If `y=x^2`
`2x^4-2x^2-40 = 0`
is equivalent to
`2y^2-2y-40 = 0`

`(2y+8)(y-5) = 0`

For zeros either
Case 1: `(2y+8) = 0`
or
Case 2: `(y-5) = 0`

Case 1: `(2y+8)=0`
`rarr y=-4`
`rarr x^2 = -4`
There are no Real solutions, but there are Complex solutions `x=+-2i`

Case 2: `(y-5)=0`
`rarr y = 5`
`rarr x^2 = 5`
`rarr x = +-sqrt(5)`