Answer
(a) $P(x)=(x+1)(x-1)(x^2+9)$
(b)$P(x)=(x+1)(x-1)(x+3i)(x-3i)$
Work Step by Step
(a) Factor P into linear and irreducible quadratic factors with real coefficients.
Assume $y=x^2$, we have $P(x)=y^2+8y-9=(y+9)(y-1)$, hence
$P(x)=(x^2+9)(x^2-1)=(x+1)(x-1)(x^2+9)$
(b) Factor P completely into linear factors with complex coefficients.
Let $x^2+9=0$, we can find the complex roots as $x=\pm 3i$, hence
$P(x)=(x+1)(x-1)(x+3i)(x-3i)$