Question

How do you use the rational root theorem to find the roots of `x^4 + 2x + 2 = 0`?

Answer 1

By the rational root theorem, any rational roots of `x^4+2x+2=0` must be `+-1` or `+-2`, neither of which works. So all the roots are irrational or complex.

Explanation:

Given `f(x) = x^4+2x+2`

By the rational root theorem, any rational factor `p/q` in lowest terms must have `p` is a divisor of the constant term `2` and `q` is a divisor of the multiplier (`1`) of the highest order term.

That means that any rational root of `f(x) = 0` must be `+-1` or `+-2`

`f(-2) = 16-4+2 = 14`
`f(-1) = 1-2+2 = 1`
`f(1) = 1+2+2 = 5`
`f(2) = 16+4+2 = 22`

So none of `+-1` or `+-2` are roots of `f(x) = 0`.
So it has no rational roots.

`f'(x) = 4x^3+2`

This is zero when `x = -1/root(3)(2)` and for no other real values of `x`. So `f(x)` has one turning point.

`f(-1/root(3)(2)) = (-1/root(3)(2))^4+2(-1/root(3)(2))+2`

`=1/(2root(3)(2))-2/root(3)(2)+2`

`=-3/2*1/root(3)(2)+2`

`> -3/2+2 = 1/2 > 0`

So `f(x) > 0` for all `x in RR`

That is `f(x) = 0` has no real roots. It has `4` complex roots.

graph{x^4+2x+2 [-11.5, 8.5, -0.8, 9.2]}