How do you find all the zeros of #f(x) = x^3 – 12x^2 + 28x – 9#?

1 Answer
May 31, 2016

Answer:

Use the rational root theorem to help find the zero #x=9#, then factor to find the other zeros: #x=3/2+-sqrt(5)/2#

Explanation:

#f(x) = x^3-12x^2+28x-9#

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 #-9# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational zeros are:

#+-1#, #+-3#, #+-9#

In addition, note that:

#f(-x) = -x^3-12x^2-23x-9#

has all negative coefficients. So #f(x)# has no negative zeros.

So the only possible rational zeros are:

#1, 3, 9#

Trying each in turn, we find:

#f(9) = 729-972+252-9 = 0#

So #x=9# is a zero and #(x-9)# a factor of #f(x)#:

#x^3-12x^2+28x-9 = (x-9)(x^2-3x+1)#

We can find the remaining two zeros using the quadratic formula:

#x = (3+-sqrt((-3)^2-4(1)(1)))/(2*1)#

#=(3+-sqrt(9-4))/2#

#=3/2+-sqrt(5)/2#