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

##### 1 Answer

#### Answer:

Use the rational root theorem to help find zero

#### Explanation:

By the rational root theorem, the only possible rational zeros of

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

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

Solve the remaining quadratic factor using the quadratic formula:

These two zeros are quite interesting in that they are powers of the golden ratio

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+...))))#