Answer
$(x-1)(x+1)(x-2i)(x+2i)$
Zeros: $-1, 1, 2i, -2i$
Work Step by Step
The question asks for the real zeros and complete factorization of P(x)
Given $P(x) = x^4 + 3x^2 - 4$
$P(x) = (x^2 + 4) (x^2 - 1)$
$P(x) = (x^2 + 4) (x-1) (x+1)$
Set each polynomial equal to zero
$x - 1 = 0$
$x = 1$
$x+1 = 0$
$x = -1$
$x^2 + 4 = 0$
$x^2 = -4$
$x = +/- 2i$
Thus the complete factorization is $(x-1)(x+1)(x-2i)(x+2i)$
The zeros are $-1, 1, 2i, -2i$
