# What are all the possible rational zeros for f(x)=2x^3-5x^2+4x-1 and how do you find all zeros?

Oct 1, 2016

So, all the zeroes of $f$ are $1 , 1 , \frac{1}{2}$.

#### Explanation:

We observe that, the sum of the co-effs. of

$f \text{ } i s = 2 - 5 + 4 - 1 = 0.$

We conclude that $\left(x - 1\right)$ is a factor of $f \left(x\right)$. Now,

$f \left(x\right) = 2 {x}^{3} - 5 {x}^{2} + 4 x - 1$

$= \underline{2 {x}^{3} - 2 {x}^{2}} - \underline{3 {x}^{2} + 3 x} + \underline{x - 1}$

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

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

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

$= \left(x - 1\right) \left\{2 x \left(x - 1\right) - 1 \left(x - 1\right)\right\}$

$= \left(x - 1\right) \left\{\left(x - 1\right) \left(2 x - 1\right)\right\}$

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

So, all the zeroes of $f$ are $1 , 1 , \frac{1}{2}$.