How to reduce a polynomial to $2$ real polynomials?

Question

Can somebody explain me the principle and to show it on this example how to find $2$ real polynomials that their multiply is $p(x)$: $$p(x)=x^{4}+1$$

Answer 1

$$x^4+1=(x^2+i)(x^2-i)=\left(x-\frac{1+i}{\sqrt2}\right)\left(x-\frac{-1-i}{\sqrt2}\right) \left(x-\frac{-1+i}{\sqrt2}\right)\left(x-\frac{1-i}{\sqrt2}\right) $$

You can see that there are two conjugate pairs (they must be so, since the overall product is a real polynomial), i.e. $\left(x-\frac{1+i}{\sqrt2}\right),\left(x-\frac{1-i}{\sqrt2}\right)$ and $\left(x-\frac{-1-i}{\sqrt2}\right),\left(x-\frac{-1+i}{\sqrt2}\right)$. You then multiply the pair to get $(x^2-\sqrt2 x+1)$ and $(x^2+\sqrt2 x+1)$. So $$ x^4+1=(x^2-\sqrt2 x+1)(x^2+\sqrt2 x+1) $$

In conclusion, you $1^{st}$ factorize a polynomial into its complex monomial, then multiply each complex conjugate pair to obtain real polynomial with degree at most 2. This work for any polynomial with real coefficients.

Answer 2

$$x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(\sqrt2 x)^2=(x^2+1+\sqrt2 x)(x^2+1-\sqrt2 x)$$

Answer 3

We know that the $4$ roots of $-1$ are $e^{\pm i\pi/4}$ and $e^{\pm i3\pi/4}$. Then, we can factor $x^4+1$ as

$$x^4+1=(x-e^{i\pi/4})(x-e^{-i\pi/4})(x-e^{i3\pi/4})(x-e^{-i3\pi/4})$$

Finally, multiplying the first and second terms and multiplying the third and fourth terms reveals that

$$\bbox[5px,border:2px solid #C0A000]{x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}$$