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

Answer:

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]}