Question

What are the possible rational roots of `2x^4-x^3-6x+3=0` and then determine the rational roots?

Answer 1

The "possible" rational roots are:

`+-1/2, +-1, +-3/2, +-3`

The only actual rational root is:

`1/2`

The other roots are the three cube roots of `3`.

Explanation:

Given:

`2x^4-x^3-6x+3 = 0`

By the rational roots theorem, any rational zeros of this quartic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `6` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational roots are:

`+-1/2, +-1, +-3/2, +-3`

Note that the ratio of the first and second terms is the same as that between the third and fourth terms. So this quartic will factor by grouping:

`0 = 2x^4-x^3-6x+3`

`color(white)(0) = (2x^4-x^3)-(6x-3)`

`color(white)(0) = x^3(2x-1)-3(2x-1)`

`color(white)(0) = (x^3-3)(2x-1)`

Hence the only rational zero is `x=1/2` and the remaining three zeros are cube roots of `3`:

`x = root(3)(3)`

`x = omega root(3)(3)`

`x = omega^2 root(3)(3)`

where `omega = -1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.