By the Rational Root Test, how many possible rational roots does #f(x)=4x^3-x^2+10x-6# have?

1 Answer
Apr 15, 2018

The equation #4x^3-x^2+10x-6 = 0# has sixteen possible rational roots (eight positive and eight negative).

Explanation:

The rational roots theorem states that if

#x = p/q#

is a rational root of the polynomial

#a_0x^n+a_1x^(n-1) +...+a_(n-1)x+a_n#

where #p,q# are mutually prime and #a_k in NN# then #p# must divide #a_n# and #q# must divide #a_0#.

For the equation:

#4x^3-x^2+10x-6 = 0#

consindering that:

#a_0 = 4 = 2 xx 2#

#a_n = -6 = -2 xx 3#

the only possible values of #q# are #+-1,+-2,+-4# and the only possible values of #p# are #+-1,+-2,+-3,+-6#.

Excluding the combinations where #p# and #q# are not mutually prime, the possibilities are then:

#+-1, +-1/2,+-1/4, +-2,+-3,+-3/2+-3/4,+-6#