How do you use the rational root theorem to find the roots of #x^4 + x^3 -2x^2 + 0x -290 = 0#?
1 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
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
Note also that:
#(color(blue)(2))^4+(color(blue)(2))^3+2(color(blue)(2))^2 = 32 < 290#
Hence there are no zeros with
So the given quartic has no rational roots.
It has one negative irrational root in
graph{x^4+x^3-2x^2-290 [-10, 10, -500, 500]}