Question

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

Answer 1

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+...))))`