What is the simplest way to factor the following polynomial $x^4-8x^2+x+12$? 
What is the simplest  way  to factor the following  polynomial  $$x^4-8x^2+x+12$$ ?

Note : I already knew the factorization which is 
$$(x^2-x-3)(x^2+x-4)$$ 
I need the way to get that 
Thank you for your help.
 A: We can write $$x^4-8x^2+x+12=(x^2+ax+b)(x^2+cx+d)$$
Comparing the coefficients of $x^3,$ $$0=a+c\iff c=-a$$
Comparing the coefficients of $x,$  $$1=ad+bc=a(d-b)$$
$$\implies a=d-b=\pm1$$
Comparing the constants,  $$bd=12$$
If $d-b=1, d=4,b=3$ or $d=-3.b=-4$
What if $d-b=-1?$
A: I like the following way.
$$x^4-8x^2+x+12=(x^2+k)^2-((2k+8)x^2-x+k^2-12).$$
Now, we need to find a value of $k$, for which $(2k+8)x^2-x+k^2-12=(ax+b)^2$,
for which we need $$1-4(2k+8)(k^2-12)=0$$ or
$$(2k+7)(4k^2+2k-55)=0.$$
Easy to see that $k=-3.5$ is valid.
Indeed, $$x^4-8x^2+x+12=(x^2-3.5)^2-(x^2-x+0.25)=$$
$$=(x^2-3.5)^2-(x-0.5)^2=(x^2-x-3)(x^2+x-4)$$
and we are done!
In the general case we'll get a cubic equation with the variable $k$ and one of this equation roots always gives difference of squares and possibility of the factorization.    
A: You can apply the algorithm described here. In your case, the resolvent of the polynomial is $x^3-16x^2+16x-1$. Clearly, $\alpha=1$ is a root of the resolvent and furthermore it is different from $0$ and it is a perfect square in $\mathbb Q$. So, you can factor your polynomial as $(x^2+x+\beta)(x^2-x+\gamma)$ (the coefficients of $x$ in these two polynomials are the square roots of $\alpha$, that is , they are $\pm1$) where $\beta$ and $\gamma$ are such that$$\left\{\begin{array}{l}\beta\gamma=12\\\beta+\gamma=\alpha+(-8)=-7.\end{array}\right.$$So, you can take $\beta=-3$ and $\gamma=-4$.
