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

2 Answers

The rational root theorem states that any rational root of a polynomial will be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
In our case #q=1# and #p=-3# hence possible roots are

#1,-1,-3,3#

But if you check none of the values above are roots for
#x^3-x^2-x-3=0#.
So the rational root theorem cannot help us here.

Dec 26, 2015

Answer:

You can't. You can only use the rational root theorem to show that it has no rational roots.

Explanation:

By the rational root theorem, any rational roots of #x^3-x^2-x-3=0# must be expressible in the form #p/q# for integers #p# and #q# with no common factor apart from #+-1#, where #p# is a factor of the constant term #-3# and #q# a factor of the coefficient #1# of the leading term.

That means that the only possible rational factors are:

#+-1#, #+-3#

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

We find:

#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 #x^3-x^2-x-3=0# has no rational roots.

In fact it has one Real root:

#x = 1/3 (1+(46-6 sqrt(57))^(1/3)+(46+6 sqrt(57))^(1/3))#

and two Complex roots:

#x = 1/3 (1+omega(46-6 sqrt(57))^(1/3)+omega^2(46+6 sqrt(57))^(1/3))#

#x = 1/3 (1+omega^2(46-6 sqrt(57))^(1/3)+omega(46+6 sqrt(57))^(1/3))#

where #omega = -1/2+sqrt(3)/2 i# is the primitive Complex cube root of #1#.

These can be found using Cardano's method or similar.