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

Sep 14, 2016

$x = 1$

#### Explanation:

$2 {x}^{3} - 3 {x}^{2} + 1 = 0$

$x = 1$ is zero. Divide $2 {x}^{3} - 3 {x}^{2} + 1$ at x-1. It will turn: $2 {x}^{2} + x + 1$

$2 {x}^{2} + x + 1 = 0$ have not rational zeros (discriminant<0)

Sep 14, 2016

The Zeroes of $f \left(x\right)$ are $1 , 1 , - \frac{1}{2}$.

#### Explanation:

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

Here, the sum of the co-effs.$= 2 - 3 + 1 = 0.$

Hence, $\left(x - 1\right)$ is a factor of $f \left(x\right)$. Rewriting $f \left(x\right)$ as,

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

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

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

In the quadr., we have the same story $\left[2 - 1 - 1 = 0\right]$ !

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

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

$= \left(x - 1\right) \left(x - 1\right) \left(2 x + 1\right)$

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

Clearly, the Zeroes of $f \left(x\right)$ are $1 , 1 , - \frac{1}{2}$.

Enjoy Maths.!