How can I simplify the polynomial $x^4+1$ into quadratic factors?

Question

The teacher gave us a hint that this polynomial expression can be written as the multiplication or sum of quadratic factors at the most. How can I do this?

Answer 1

$$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt 2x)^2=(x^2-\sqrt 2x+1)(x^2+\sqrt 2x+1).$$

Answer 2

do you mean $x^4+1$? in which case you can factor by complex polynomials i.e. $$x^4+1=(x^2-i)(x^2+i)$$

Answer 3

mathlove already gave an excellent answer but here is another, for the sake of completeness:

Let $P(x)=ax^2+bx+c$ and $Q(x)=dx^2+ex+f$ such that $P(x)Q(x)=x^4+1$. Note that, if $\lambda$ is not zero, then $\lambda P$ and $\frac{1}{\lambda}Q$ are also solutions, so I can assume that $a=1$.

Now, by expanding $PQ$, I obtain $P(x)Q(x)=dx^4+(e+bd)x^3+(f+be+cd)x^2+(bf+ce)x+cf$. This means that $d=1$, $e+bd=0$, $f+be+cd=0$, $bd+ce=0$, and $cf=1$. Solving these equarions give $d=1$, $b=\pm\sqrt{2}$, $e=-b$, $c=1$ and $f=1$ (There are two possibilities for $b$ since permuting $P$ and $Q$ gives another solution to the problem).