# What are all the possible rational zeros for f(x)=x^3-13x^2+23x-11 and how do you find all zeros?

Jan 22, 2017

The "possible" rational zeros are: $\pm 1 , \pm 11$

The actual zeros are: $1 , 1 , 11$

#### Explanation:

$f \left(x\right) = {x}^{3} - 13 {x}^{2} + 23 x - 11$

By the rational roots 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 $- 11$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 11$

Note that the sum of the coefficients of $f \left(x\right)$ is zero. That is:

$1 - 13 + 23 - 11 = 0$

Hence $f \left(1\right) = 0$, $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{3} - 13 {x}^{2} + 23 x - 11 = \left(x - 1\right) \left({x}^{2} - 12 x + 11\right)$

The sum of the coefficients of the remaining quadratic is zero too:

$1 - 12 + 11 = 0$

So $x = 1$ is a zero again and $\left(x - 1\right)$ a factor:

$\left({x}^{2} - 12 x + 11\right) = \left(x - 1\right) \left(x - 11\right)$

So the zeros of $f \left(x\right)$ are $x = 1$ with multiplicity $2$ and $x = 11$.