Factoring quartic equation

Question

$$x^4+6x^2+25=0$$

How could I factor it into $(x^2-2x+5)(x^2+2x+5)=0$?

I got the result looking into horrible formulas on wikipedia, but I suppose there's a much easier way. Could you help me?

Answer 1

$$\underbrace{(x^2)^2+5^2+2\cdot x^2\cdot5}+(6-10)x^2=(x^2+5)^2-(2x)^2$$

Answer 2

Notice that $$ x^4+6x^2+25=(x^4+6x^2+9)+16=(x+3)^2+4^2=(x^2+3+4i)(x^2+3-4i) \quad \forall x\in \mathbb{C}. $$ Let $u \in \mathbb{C}$ such that $u^2=3+4i$. It is easy to show that $u\in\{2+i,-2-i\}$. We have \begin{eqnarray} x^4+6x^2+25&=&(x^2+u^2)(x^2+\bar{u}^2)=(x+iu)(x-iu)(x+i\bar{u})(x-i\bar{u})\\ &=&(x+iu)\overline{(x+iu)}(x-iu)\overline{(x-iu)}\\ &=&|x+iu|^2|x-iu|^2=|x-1+2i|^2|x+1-2i|^2\\ &=&[(x-1)^2+4][(x+1)^2+4]=(x^2-2x+5)(x^2+2x+5). \end{eqnarray}

Answer 3

Why not simply replace $x^2$ by $y$. You then end with a quadratic equation in $y$ from which later uou could extract $x$. The solutions of the quadratic are $-3-4 i$ and $-3+4 i$. Taking the square roots, you then obtain $$x=-1 - 2 I$$ $$x=-1 + 2 I$$ $$x=1 - 2 I$$ $$x=1 - 2 I$$.