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

1 Answer
Aug 4, 2018

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]}