Answer
$P(x)=(x-1)(x+1)(x+2)(x-2)$
Zeros: $ \ -2,\ -1, \ 1, \ 2. $
Work Step by Step
$P(x)=x^{4}-5x^{2}+4$
$P(-x)=x^{4}-5x^{2}+4$
Descart's rule of signs:
P(x) has 2 sign changes $\Rightarrow$ 2 or 0 positive zeros.
P(-x) has $2$ sign changes $\Rightarrow$ 2 or 0 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 4$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 4$
Testing with synthetic division,
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 & 0 & -5 & 0 & 4\\\hline
& 1 & 1 & -4 & -4\\\hline
1& 1 & -4 & -4&|\ \ 0\end{array}$
$P(x)=(x-1)(x^{3}+x^{2}-4x-4)$
Again, testing with synthetic division,
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1& 1 & -4 & -4\\\hline
& -1 & 0 & 4\\\hline
1& 0 & -4 &|\ \ 0\end{array}$
$P(x)=(x-1)(x+1)(x^{2}-4)$
... recognize a difference of squares
$P(x)=(x-1)(x+1)(x+2)(x-2)$
Zeros: $ \ -2,\ -1, \ 1, \ 2. $
