# What are all the possible rational zeros for f(x)=4x^4-17x^2+4 and how do you find all zeros?

Oct 7, 2016

$\text{All the } 4$ possible zeroes of $f \text{ } a r e , \pm \frac{1}{2} , \pm 2.$

#### Explanation:

$f \left(x\right) = 4 {x}^{4} - 17 {x}^{2} + 4$

To complete the square $4 {x}^{4} + 4 ,$ we find the Middle Term$= 8 {x}^{2.}$

$\therefore f \left(x\right) = 4 {x}^{4} + 8 {x}^{2} + 4 - 25 {x}^{2}$

$= {\left(2 {x}^{2} + 2\right)}^{2} _ {\left(5 x\right)}^{2}$

$= \left(2 {x}^{2} + 5 x + 2\right) \left(2 {x}^{2} - 5 x + 2\right)$

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

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

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

$\therefore \text{ all the } 4$ possible zeroes of $f \text{ } a r e , \pm \frac{1}{2} , \pm 2.$