How do you use the rational roots theorem to find all possible zeros of #P(x) = x^3 - 4x^2 + x + 6#?

1 Answer
Jul 30, 2016

Answer:

Zeros: #-1, 2, 3#

Explanation:

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

By the rational roots theorem, any rational zeros of #P(x)# are expressible as #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#

We find:

#P(-1) = -1-4-1+6 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

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

We could simply evaluate #P(x)# for the remaining possible zeros, but #x^2-5x+6# factors fairly easily:

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

So the other two zeros are #x=2# and #x=3#.