Answer
(a) $ (x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)$
(b) $\frac{-\sqrt 2\pm i\sqrt {2}}{2},\frac{\sqrt 2\pm i\sqrt {2}}{2}$.
Work Step by Step
(a) $f(x)=x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)$
(b) For $x^2+\sqrt 2x+1=0$, we have $x=\frac{-\sqrt 2\pm\sqrt {2-4}}{2}=\frac{-\sqrt 2\pm i\sqrt {2}}{2}$. For $x^2-\sqrt 2x+1=0$, we have $x=\frac{\sqrt 2\pm\sqrt {2-4}}{2}=\frac{\sqrt 2\pm i\sqrt {2}}{2}$.
