How do you use the rational roots theorem to find all possible zeros of #f(x)=x^3-5x^2+2x+12#?

1 Answer
Jul 24, 2016

Answer:

#x=3# or #x=1+-sqrt(5)#

Explanation:

#f(x) = x^3-5x^2+2x+12#

By the rational root theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #12# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-6, +-12#

Trying each in turn we find:

#f(3) = 27-45+6+12 = 0#

So #x=3# is a zero and #(x-3)# a factor:

#x^3-5x^2+2x+12#

#= (x-3)(x^2-2x-4)#

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

#= (x-3)((x-1)^2-(sqrt(5))^2)#

#= (x-3)(x-1-sqrt(5))(x-1+sqrt(5))#

So the remaining two zeros are:

#x = 1+-sqrt(5)#