Factor $x^4+1$ as product of two factors

Question

I want to do partial fraction expansion on $$\frac{1}{x^4+1}.$$ But how to I factor the denominator so that I get a product of two factors?

Answer 1

hint

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

$=(x^2+1+x\sqrt{2})(x^2+1-x\sqrt{2})$

Answer 2

The roots of $x^4+1$ are complex, pairwise conjugate, namely $\dfrac{\pm1\pm i}{\sqrt2}$ (they are the square roots of $\pm i$). Then developing and using Vieta's formulas,

$$(x-r_{++})(x-r_{+-})(x-r_{-+})(x-r_{--})=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1).$$

Answer 3

note that $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt{2}x)^2=(x^2+1-\sqrt{2}x)(x^2+1+\sqrt{2}x)$$

Answer 4

Trying $(x^2+ax+1)(x^2-ax+1) =x^4+x^2(2-a^2)+1 $, we find that $a^2=2$ works.

Answer 5

Every quartic polynomial factors as the product of two quadratic ones. As both the coefficients of $x^4$ and $x^0$ are $1$, the factorization must be of the form

$$(x^2+ax+1)(x^2+bx+1).$$

Developing, you get

$$x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1.$$

Clearly, $a=-b=\pm\sqrt2$.