Question

For the polynomial equation `3x^3-x^2- 7x+8`, how do you determine all possible rational roots?

Answer 1

The only "possible" rational zeros are:

`+-1/3, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8`

Explanation:

Given:

`3x^3-x^2-7x+8`

By the rational root theorem, any rational zeros of this cubic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `8` and `q` a divisor of the coefficient `3` of the leading term.

That means that the only possible rational zeros are:

`+-1/3, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8`

In fact, none of these work, so this cubic only has irrational and/or complex zeros. Its only real zero is approximately `-1.79`

graph{3x^3-x^2-7x+8 [-5, 5, -2.52, 15]}