# How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given f(x)=x^4-5x^2+4?

Mar 5, 2017

 (1) : f(x)" has two +ve zeroes, namely, "2 and 1;

$\left(2\right) : \text{ two "-ve "zeroes, } - 2 , \mathmr{and} , - 1 :$

$\left(3\right) : \text{ four rational zeroes, } \pm 2 , \pm 1$.

#### Explanation:

Let ${x}^{2} = y$ in $f \left(x\right) = {x}^{4} - 5 {x}^{2} + 4 = {y}^{2} - 5 y + 4.$

$\therefore f \left(x\right) = 0 \Rightarrow {y}^{2} - 5 y + 4 = 0.$

$\therefore \underline{{y}^{2} - 4 y} - \underline{y + 4} = 0.$

$\therefore y \left(y - 4\right) - 1 \left(y - 4\right) = 0.$

$\therefore \left(y - 4\right) \left(y - 1\right) = 0.$

$\therefore \left({x}^{2} - 4\right) \left({x}^{2} - 1\right) = 0 , \ldots \ldots \ldots . . \left[\because , y = {x}^{2}\right]$

$\therefore \left(x + 2\right) \left(x - 2\right) \left(x + 1\right) \left(x - 1\right) = 0.$

$\therefore x = - 2 , 2 , - 1 , \mathmr{and} , 1.$

Thus,  (1) : f(x)" has two +ve zeroes, namely, "2 and 1;

$\left(2\right) : \text{ two "-ve "zeroes, } - 2 , \mathmr{and} , - 1 :$

$\left(3\right) : \text{ four rational zeroes, } \pm 2 , \pm 1$.

Enjoy Maths.!