What are the zeroes of # f(x) = x^3 – 12x^2 + 28x – 9#?

1 Answer
Dec 23, 2015

Answer:

Use the rational root theorem to help find zero #x=9#, then divide #f(x)# by #(x-9)# to find the quadratic #x^2-3x+1# which we can solve using the quadratic formula to find the irrational zeros.

Explanation:

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

By the rational root theorem, the only possible rational zeros of #f(x)# can be expressed in lowest terms in the form #p/q# where #p# and #q# are integers with no common factors larger than #1#, #p# is 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#

Trying each of these in turn, we eventually find:

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

#=729-(12*81)+(28*9)-9#

#=729-972+252-9 = 0#

So #(x-9)# is a factor of #f(x)# ...

#x^3-12x^2+28x-9#

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

Solve the remaining quadratic factor using the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (3+-sqrt(3^2-(4xx1xx1)))/2 = (3+-sqrt(5))/2#

These two zeros are quite interesting in that they are powers of the golden ratio #phi = (1+sqrt(5))/2#, namely #phi^2# and #phi^-2#.

As a result they have simple continued fraction expansions:

#(3+sqrt(5))/2 = phi^2 = [2;bar(1)] = 2 + 1/(1+1/(1+1/(1+...)))#

#(3-sqrt(5))/2 = phi^-2 = [0;2,bar(1)] = 0 + 1/(2+1/(1+1/(1+1/(1+...))))#