Question

How do you use the rational root theorem to find the roots of `f(x)=x^4-x-4`?

Answer 1

You can't. `f(x) = 0` has no rational roots.

Explanation:

By the rational root theorem, any rational root of `x^4-x-4 = 0` is expressible in lowest terms as `p/q` where `p, q in ZZ`, `q != 0`, `p` a divisor of the constant term `4` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational roots are:

`+-1`, `+-2`, `+-4`

Then we find:

`f(-4) = 256+4-4 = 256`
`f(-2) = 16+2-4 = 14`
`f(-1) = 1+1-4 = -2`
`f(1) = 1-1-4 = -4`
`f(2) = 16-2-4 = 10`
`f(4) = 256-4-4 = 248`

So `f(x) = 0` has no rational roots.

Since `f(x)` is continuous, we can tell that `f(x) = 0` has Real zeros in `(-2, -1)` and `(1, 2)`.

Newton's method can then be used to find approximations for the two Real roots. Starting with an approximation `a_0`, find better approximations by iterating using the formula:

`a_(i+1) = a_i - f(x)/(f'(x))`

`f'(x) = 4x^3-1`

Starting with `a_0 = -1.5`, we find:

`a_1 = -1.3232758621`
`a_2 = -1.2853460646`
`a_3 = -1.2837842190`
`a_4 = -1.2837816659`
`a_5 = -1.2837816659`

Starting with `a_0 = 1.5`, we find:

`a_1 = 1.535`
`a_2 = 1.5337528051`
`a_3 = 1.5337511688`
`a_4 = 1.5337511688`

graph{x^4-x-4 [-10, 10, -5, 5]}