Answer
The zeros of $f\left( x \right)={{x}^{4}}-5{{x}^{2}}+4$ are $1,-1,2,-2$.
Work Step by Step
Let’s first equate $f\left( x \right)$ to $0$ as below. Then,
$\begin{align}
& {{x}^{4}}-5{{x}^{2}}+4=0 \\
& {{x}^{4}}-4{{x}^{2}}-{{x}^{2}}+4=0 \\
& {{x}^{2}}\left( {{x}^{2}}-4 \right)-\left( {{x}^{2}}-4 \right)=0 \\
& \left( {{x}^{2}}-4 \right)\left( {{x}^{2}}-1 \right)=0
\end{align}$
Now, by using the formula $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ , the polynomial can be written as:
$\left( x-1 \right)\left( x+1 \right)\left( x-2 \right)\left( x+2 \right)=0$
Put each factor equal to $0$. So,
$\left( x-1 \right)=0$
Or,
$\left( x+1 \right)=0$
Or,
$\left( x-2 \right)=0$
Or,
$\left( x+2 \right)=0$
The above equations provide the values of x as:
$x=1,-1,2,-2$
