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

May 31, 2016

Use the rational root theorem to help find the zero $x = 9$, then factor to find the other zeros: $x = \frac{3}{2} \pm \frac{\sqrt{5}}{2}$

#### Explanation:

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

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ 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:

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

$f \left(- x\right) = - {x}^{3} - 12 {x}^{2} - 23 x - 9$

has all negative coefficients. So $f \left(x\right)$ has no negative zeros.

So the only possible rational zeros are:

$1 , 3 , 9$

Trying each in turn, we find:

$f \left(9\right) = 729 - 972 + 252 - 9 = 0$

So $x = 9$ is a zero and $\left(x - 9\right)$ a factor of $f \left(x\right)$:

${x}^{3} - 12 {x}^{2} + 28 x - 9 = \left(x - 9\right) \left({x}^{2} - 3 x + 1\right)$

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

$x = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(1\right)}}{2 \cdot 1}$

$= \frac{3 \pm \sqrt{9 - 4}}{2}$

$= \frac{3}{2} \pm \frac{\sqrt{5}}{2}$