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

1 Answer
Nov 10, 2015

Answer:

The theorem states that the polynomial has no rational roots.

Explanation:

The rational root theorem states that if you have a polynomial with integer coefficients (as you have), then a solution #x=p/q#, where #p# and #q# have no common divisors, must be such that #p# divides #a_0#, and #q# divides the leading coefficient.

Since in your case the leading coefficient is the coefficient of #x^3#, which is one, then #q# must divide one, and so it must be one itself. On the other hand, #p# must divide #-3#, which means that it can be #-3,-1,1,3#.

So, we can check these four values, because we know what if there's a solution, it must be one of these four.

Let #p(x)=x^3-x^2-x-3#. Then,

  • #p(-3)=(-3)^3 - (-3^2) - (-3) -3 = #

#-27 - 9 - (-3) -3 = #

#-27-9+3-3 = -36 ne 0#

  • #p(-1)=(-1)^3 - (-1^2) - (-1) -3 = -1-1+1-3 = -4 ne 0#
  • #p(1)=1^3 -1^2 - 1 -3 = 1-1-1-3 = -4 ne 0 #
  • #p(3)=3^3 -3^2 - 3 -3 = 27-9-3-3 = 12 ne 0 #

This means that the polynomial has no rational roots.