Question

Find the zeros of f(x) = 16x^4 - 81?

Answer 1

So, if f(x)=0=16x^4-81, 0=(4x^2+9)(4x^2-9)=(4x^2+9)(2x+3)(2x-3), thus, the x's are as Mr. Rivera put it. There is also a complex solution of 4x^2=-9.

Answer 2

the complex solution would be \[\sqrt{3/2} \] i on the outside?

Answer 3

Yes!