How to factorize $x^4+2x^2+4$ to a product of polynomials with real coefficients?

Question

How do you factor

$$x^4+2x^2+4 $$

so it can be written as

$$ (x^2+2x+2)(x^2-2x+2) $$

Answer 1

I'm thinking you have a slight mistake in there. In any case, you can rewrite this as

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

from which point it can be treated as the difference of $2$ squares to obtain

$$(x^2+x\sqrt2+2)(x^2-x\sqrt2+2)$$

Answer 2

The key idea is to complete $\,x^4\!+4\,$ to a square $\,(x^2\!+2)^2-4x^2,\,$ yielding a difference of squares

$$ x^4\!+4\, =\, (x^2\!+2)^2 - (\color{#c00}{2x})^2 =\, (x^2\!+2-\color{#c00}{2x})(x^2\!+2 + \color{#c00}{2 x})$$

The same idea works for $\,x^4+4 + 2x^2\,$ if you meant that instead of $\,x^4+4.$

Answer 3

Certainly the solution using completing the square (already provided in two answers) is in this case the easiest one.

I will just point out that in general if you have a real polynomial which is simple enough so that we are able to find all complex roots, this might also help you find factorization over $\mathbb R$. This works because for every root $z=a+bi$ also the complex conjugate $\overline z=a-bi$ is a root. And these pairs give you quadratic factors $$(x-z)(x-\overline z)=(x-a-bi)(x-a+bi)=x^2-2a+b^2+b^2.$$ See also: Complex conjugate root theorem

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In this case you can use substitution $t=x^2$ to get a quadratic equation $$t^2+2t+4=0$$ which has the solutions $t_{1,2}=\frac{-2\pm\sqrt{12}i}2=-1\pm\sqrt3i = 2\left(-\frac12\pm\frac{\sqrt3}2i\right)$.

Now you can solve the equations $$x^2=-1\pm\sqrt3i$$ which have the solutions \begin{align*} x_1&=\sqrt2\left(\frac12+\frac{\sqrt3}2i\right)\\ x_2&=\sqrt2\left(-\frac12\pm\frac{\sqrt3}2i\right)\\ x_3&=\overline{x_1}\\ x_4&=\overline{x_2} \end{align*} This will give you the following quadratic factors: \begin{align*} (x-x_1)(x-\overline{x_1}) = x^2-\sqrt2x+2\\ (x-x_2)(x-\overline{x_2}) = x^2+\sqrt2x+2 \end{align*}