Question

How do you find all the roots of `f(x) = x^4 + 2x^3 + x^2 - 2x - 2`?

Answer 1

This quartic polynomial has zeros `+-1` and `-1+-i`.

Explanation:

First note that the sum of the coefficients is `0`. That is, `1+2+1-2-2 = 0`. Hence we can deduce that `x=1` is a zero and `(x-1)` a factor:

`x^4+2x^3+x^2-2x-2 = (x-1)(x^3+3x^2+4x+2)`

If you reverse the signs of the coefficients of the terms of odd degree in the remaining cubic `(x^3+3x^2+4x+2)` you get: `-1+3-4+2 = 0`. Hence `x=-1` is a zero and `(x+1)` a factor:

`x^3+3x^2+4x+2 = (x+1)(x^2+2x+2)`

The remaining quadratic factor has negative discriminant, but you can factor it by completing the square with Complex coefficients:

`x^2+2x+2 = x^2+2x+1+1 = (x+1)^2-i^2 = (x+1-i)(x+1+i)`

Hence zeros `x=-1+-i`