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

1 Answer
Jul 4, 2018

Answer:

The "possible" rational zeros are #+-1, +-2, +-3, +-6#.

The only real zero is #x=1#

Explanation:

Given:

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

By the rational roots theorem any rational zeros of #P(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 #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-6#

In addition, note that the sum of the coefficients is zero, i.e.:

#1+3+2-6 = 0#

Hence we can deduce that #x=1# is a zero and #(x-1)# a factor:

#x^5+3x^3+2x-6 = (x-1)(x^4+x^3+4x^2+4x+6)#

The remaining quartic has all positive coefficients, so any real zeros must be negative.

Trying #-1#, #-2#, #-3# and #-6# we find that none is a zero.

In fact it has only non-real complex zeros.