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

Dec 23, 2015

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

#### Explanation:

$f \left(x\right) = {x}^{3} - 12 {x}^{2} + 28 x - 9$

By the rational root theorem, the only possible rational zeros of $f \left(x\right)$ can be expressed in lowest terms in the form $\frac{p}{q}$ 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:

$\pm 1$, $\pm 3$, $\pm 9$

Trying each of these in turn, we eventually find:

$f \left(9\right) = {9}^{3} - \left(12 \cdot {9}^{2}\right) + \left(28 \cdot 9\right) - 9$

$= 729 - \left(12 \cdot 81\right) + \left(28 \cdot 9\right) - 9$

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

So $\left(x - 9\right)$ is a factor of $f \left(x\right)$ ...

${x}^{3} - 12 {x}^{2} + 28 x - 9$

$= \left(x - 9\right) \left({x}^{2} - 3 x + 1\right)$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{3 \pm \sqrt{{3}^{2} - \left(4 \times 1 \times 1\right)}}{2} = \frac{3 \pm \sqrt{5}}{2}$
These two zeros are quite interesting in that they are powers of the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$, namely ${\phi}^{2}$ and ${\phi}^{-} 2$.
(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+...))))