For the polynomial equation #3x^3-x^2- 7x+8#, how do you determine all possible rational roots?

1 Answer
Mar 16, 2017

Answer:

The only "possible" rational zeros are:

#+-1/3, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8#

Explanation:

Given:

#3x^3-x^2-7x+8#

By the rational root theorem, any rational zeros of this cubic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #8# and #q# a divisor of the coefficient #3# of the leading term.

That means that the only possible rational zeros are:

#+-1/3, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8#

In fact, none of these work, so this cubic only has irrational and/or complex zeros. Its only real zero is approximately #-1.79#

graph{3x^3-x^2-7x+8 [-5, 5, -2.52, 15]}