Question

How do you find the complex roots of `4m^4+17m^2+4=0`?

Answer 1

The roots are:

`m = +-1/2i" "` and `" "m = +-2i`

Explanation:

`4m^4+17m^2+4 = 0`

Treating this as a quadratic in `m^2`, we can use an AC method to factor it into quadratics, then use the difference of squares identity with Complex coefficients.

First find a pair of factors of `AC = 4*4 = 16` with sum `B=17`.

The pair `16, 1` works.

Use this pair to split the middle term, then factor by grouping:

`4m^4+17m^2+4 = 4m^4+16m^2+m^2+4`

`color(white)(4m^4+17m^2+4) = (4m^4+16m^2)+(m^2+4)`

`color(white)(4m^4+17m^2+4) = 4m^2(m^2+4)+1(m^2+4)`

`color(white)(4m^4+17m^2+4) = (4m^2+1)(m^2+4)`

`color(white)(4m^4+17m^2+4) = ((2m)^2-i^2)(m^2-(2i)^2)`

`color(white)(4m^4+17m^2+4) = (2m-i)(2m+i)(m-2i)(m+2i)`

Hence zeros:

`m = +-1/2i" "` and `" "m = +-2i`