How do you use the rational roots theorem to find all possible zeros of #3x^5-7x^2+x+6#?

1 Answer
May 24, 2016

Answer:

This quintic has no rational zeros.

Explanation:

#f(x) = 3x^5-7x^2+x+6#

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #6# 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#, #+-2#, #+-3#, #+-6#

Evaluating #f(x)# for each of these, we find none work.

So this quintic has no rational zeros.

In fact it has one Real zero at #x~~ -0.873# and two pairs of Complex zeros.

These zeros are not expressible in terms of #n#th roots, so about the best you can do is find numerical approximations using Newton's method or the Durand-Kerner method.