Answer
$P(x)=(x^2+4)(x-1)(x+1)$
Work Step by Step
$P(x)=x^4+3x^2-4$.
To factorise the trinomial,
lett $x^2=k$,
$k^2+3k-4$,
factorze the trinomial $k^2+3k-4$,
(find factors of $-4(1)=-4$ whose sum is $3$):
($-1$ and $+4$),
$k^2-1k+4k-4=k(k-1)+4(k-1)=(k+4)(k-1)$,
Lets replace $x^2=k$, into the factors
$P(x)=(x^2+4)(x^2-1)$,
$P(x)=(x^2+4)(x-1)(x+1)$
The binomial $x^2+4$ cannot be further decomposed with real coefficients.
