# What are all the possible rational zeros for f(x)=2x^3-4x+8 and how do you find all zeros?

Sep 20, 2016

Thus, all the zeroes of $f$ are, $- 2 , 1 + i , \mathmr{and} , 1 - i$; of these, only

$- 2$ is rational.

#### Explanation:

We have, $f \left(x\right) = 2 {x}^{3} - 4 x + 8 = 2 \left({x}^{3} - 2 x + 4\right) = 2 g \left(x\right) , s a y , w h e r e , g \left(x\right) = {x}^{3} - 2 x + 4$.

Clearly, $\left(x \pm 1\right)$ are not the factors of $g \left(x\right)$.

$\text{Now, the Leading Co-eff. of "g" is "1", and the Const. Term is } 4$,

( factors $1 , 2 , 4$), we can guess the probable factors of $g$ as

(x+-1), (x+-2), &, (x+-4).

We have already checked that $\left(x \pm 1\right)$ are not factors.

If $\left(x - 2\right) | g \left(x\right) \text{, then, "g(2)" must be 0, but, "g(2)=8-4+4ne0, so, (x-2)" is not a factor.}$

$g \left(- 2\right) = - 8 + 4 + 4 = 0 \Rightarrow \left(x + 2\right) \text{ is a factor of } g \left(x\right)$.

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

$= \underline{{x}^{3} + 2 {x}^{2}} - \underline{2 {x}^{2} - 4 x} + \underline{2 x + 4}$

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

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

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

${x}^{2} - 2 x + 2 , \Delta = {\left(- 2\right)}^{2} - 4 \left(1\right) \left(2\right) = 4 - 8 = - 4 < 0$.

Hence, that quadr. can not have rational zeroes. Its complex

zeroes are $\frac{2 \pm 2 i}{2} = \left(1 \pm i\right)$

Thus, all the zeroes of $f$ are, $- 2 , 1 + i , \mathmr{and} , 1 - i$; of these, only

$- 2$ is rational.

Enjoy Maths!