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

1 Answer
George C.
Sep 9, 2015

The rational root theorem only helps us find that #x^3-x^2-x-3 = 0# has no rational roots.

The only Real root is:

#x = 1/3+root(3)(46+6sqrt(57))/3+root(3)(46-6sqrt(57))/3 ~~ 2.1304#

Explanation:

Let #f(x) = x^3-x^2-x-3#.

By the rational root theorem, any rational roots of #f(x) = 0# must be of the form #p/q# in lowest terms, where #p, q in ZZ#, #q != 0#, #p# is a divisor of the constant term #-3# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational roots of #f(x) = 0# are:
#+-1#, #+-3#

Let's try them:

#f(1) = 1 - 1 - 1 -3 = -4#
#f(-1) = -1-1+1-3 = -4#
#f(3) = 27-9-3-3 = 12#
#f(-3) = -27-9+3-3 = -36#

So #f(x) = 0# has no rational roots and the rational root theorem cannot help us find the roots that do exist.

Using a simple Tschirnhaus transformation (#t = x - 1/3#) and Cardano's method, I found:

#x = 1/3+root(3)(46+6sqrt(57))/3+root(3)(46-6sqrt(57))/3 ~~ 2.1304#

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