Answer
(a) $P(x)=(x^2+4)(x^2+4)$
(b) $P(x)=(x+2i)(x+2i)(x-2i)(x-2i)$
Work Step by Step
(a) Factor P into linear and irreducible quadratic factors with real coefficients.
Assume $y=x^2$, we have $P(x)=y^2+8y+16=(y+4)^2$, hence
$P(x)=(x^2+4)^2=(x^2+4)(x^2+4)$
(b) Factor P completely into linear factors with complex coefficients.
Let $x^2+4=0$, we can find the complex roots as $x=\pm 2i$, hence
$P(x)=(x+2i)^2(x-2i)^2=(x+2i)(x+2i)(x-2i)(x-2i)$