Question

How do you find all the real and complex roots of `x^4 - 2x^3 - 4x^2 - 8x - 32 = 0`?

Answer 1

Use the rational root theorem, then divide by the factors found to find the remaining roots.

`x=-2`, `x=4`, `x=+-2i`

Explanation:

Let `f(x) = x^4-2x^3-4x^2-8x-32`

By the rational root theorem, the only possible rational roots of `f(x) = 0` are expressible in the form `p/q` for some integers `p` and `q` where `p` is a divisor of the constant term `-32` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots are factors of `32`:

`+-1`, `+-2`, `+-4`, `+-8`, `+-16`, `+-32`

Trying each of these in turn, we find:

`f(-2) = 16+16-16+16-32 = 0`

`f(4) = 256-128-64-32-32 = 0`

So `x=-2` and `x=4` are roots and `f(x)` has corresponding factors `(x+2)` and `(x-4)`

`x^4-2x^3-4x^2-8x-32`

`= (x+2)(x^3-4x^2+4x-16)`

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

`= (x+2)(x-4)(x-2i)(x+2i)`