Question

How do you use the rational root theorem to find the roots of `x^4 + x^3 -2x^2 + 0x -290 = 0`?

Answer 1

The "possible" rational roots are:

`+-1, +-2, +-5, +-10, +-29, +-58, +-145, +-290`

but this quartic has no rational roots.

Explanation:

Given:

`x^4+x^3-2x^2+0x-290 = 0`

Assuming there's no typographical error in the question, this can be written:

`x^4+x^3-2x^2-290 = 0`

By the rational roots theorem, any rational roots of this quartic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-290` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots are:

`+-1, +-2, +-5, +-10, +-29, +-58, +-145, +-290`

In addition, note that:

`(color(blue)(5))^4 = 629 > 465 = (color(blue)(5))^3+2(color(blue)(5))^2+290`

Hence there are no zeros with `abs(x) >= 5`.

Note also that:

`(color(blue)(2))^4+(color(blue)(2))^3+2(color(blue)(2))^2 = 32 < 290`

Hence there are no zeros with `abs(x) <= 2`.

So the given quartic has no rational roots.

It has one negative irrational root in `(-5, -2)`, one positive irrational root in `(2, 5)` and two non-real complex roots.

graph{x^4+x^3-2x^2-290 [-10, 10, -500, 500]}