# What are all the possible rational zeros for f(x)=4x^3-9x^2+6x-1 and how do you find all zeros?

Feb 19, 2017

$f \left(x\right)$ has "possible" zeros:

$\pm \frac{1}{4}$, $\pm \frac{1}{2}$, $\pm 1$

and actual zeros:

$\frac{1}{4} , 1 , 1$

#### Explanation:

Given:

$f \left(x\right) = 4 {x}^{3} - 9 {x}^{2} + 6 x - 1$

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 $- 1$ and $q$ a divisor of the coefficient $4$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm 1$

Here's a non-standard way of finding the zeros:

I notice that the coefficients are all fairly small and the signs alternate.

Taking them in reverse order and stringing them together we get:

$1694 = 2 \cdot 7 \cdot 11 \cdot 11 = 14 \cdot 11 \cdot 11$

Hence we find:

${t}^{3} + 6 {t}^{2} + 9 t + 4 = \left(t + 4\right) \left(t + 1\right) \left(t + 1\right)$

Then putting $t = - \frac{1}{x}$ we find:

$4 {x}^{3} - 9 {x}^{2} + 6 x - 1 = \left(4 x - 1\right) \left(x - 1\right) \left(x - 1\right)$

So $f \left(x\right)$ has zeros $\frac{1}{4} , 1 , 1$