Factorize $x^4+16x-12$ over reals Factorize $x^4+16x-12$ over reals.
The factor is $x^4+16x-12=(x^2-2x+6)(x^2+2x-2)$
It can be factorized again but I am stuck in this step.If we want to add and then subtract we have a lot of thing to add and subtract.Another idea that I saw in books  is writing as this:
$x^4+16x-12=(x^2+ax+b)(x^2+a'x+b')$
and then find $a,b,b',a'$ but there are two problems I can't find these here and we can say maybe it can factorized into one degree $3$ and one degree $1$ polynomial.
Isn't there a nice way to factor this?
 A: You can check that it's not linear-times-cubic by looking for rational roots, which must have the form $p/q$ where $p$ divides 12 and $q$ divides 1. When that doesn't work, the product-of-quadratics approach is your next best bet. 
A: @JohnHughes explains how to find that this is not a cubic times linear factor. Once you figure that out, we know it's a quadratic times quadratic. To solve this, I'm going to go off the equation you had:
$$x^4+16x-12=(x^2+ax+b)(x^2+a'x+b')$$
Expand the right side:
$$x^4+16x-12=x^4+(a+a')x^3+(b+b'+aa')x^2+(ab'+ba')x+bb'$$
Now, we have $0x^3$ on the left side and $(a+a')x^3$ on the right side, so we have $a+a'=0 \implies a'=-a$. Furthermore, we have $-12$ as our constant on the left side and $bb'$ as our constant on the right side, so we have $bb'=-12 \implies b'=-\frac{12}b$. Susbtitute:
$$x^4+16x-12=x^4+\left(b-\frac{12}{b}-a^2\right)x^2+\left(-\frac{12a}{b}-ab\right)x-12$$
We have $0x^2$ and $+16x$ on the left side, so by comparing such with the $x^2$ and $x$ coefficients on the right side, we get the following equations:
$$b-\frac{12}{b}-a^2=0$$
$$-\frac{12a}{b}-ab=16$$
Multiply both sides by both equations $b$:
$$b^2-12-ba^2=0$$
$$-12a-ab^2=16b$$
However, instead of guess and check, we can solve for $a$ in terms of $b$ in the second equation (I choose the second equation because it is easier):
$$a=\frac{-16b}{12+b^2}$$
Now, remember that $b$ is a factor of $12$ since $bb'=-12$, so $b \in \{-12,-6,-4,-3,-2,-1,1,2,3,4,6,12\}$. Guess and check values of $b$ and solve for $a$. When you get an integer value of $a$, you know you've solved the problem, so substitute back into $a+a'=0$ to find $a'$ and $bb'=-12$ to find $b'$.
A: $$x^4+16x-12=(x^4+4x^2+4)-(4x^2-16x+16)$$
A: What you want is to show that the resolvent cubic associated with the quartic has a rational root.  Look at this method based on the usual "general solution" for quartics:
Assume a factorization if the form:
$x^4+px^2+qx+r=$
$(x^2-2\sqrt{s}x+t_1)(x^2+2\sqrt{s}x+t_2)$
Then expand the right side and match terms of like degree:
Degree 4:  $1=1$ (forced)
Degree 3:  $0=0$ (forced)
Degree 2:  $p=-4s+(t_1+t_2)$
Degree 1:  $q=(2\sqrt{s})(t_1-t_2)$
Degree 0:  $r=t_1t_2$
We may treat the degree 2 and degree 1 equations as a linear system for $t_1, t_2$ in terms of $s$, then shbstitute the solution of this system into the degree 0 equation.  Now that equation contains only $s$ as unknown, and after the usual simplifications we get this resolvent cubic:
$s^3-(p/2)s^2+((p^2-4r)/16)s-(q^2/64)=0$
This cubic equation has three roots corresponding to three ways the quartic roots could be paired to make quadratic factors.  If you can find a rational root of this resolvent cubic you get a simplified quadratic factorization.
Here, the resolvent cubic is
$s^3+3s-4=0$
with the rational root $s=1$.  Then we have a factorization
$(x^2-2x+t_1)(x^2+2x+t_2)$
where $t_1,t_2$ are determined as above with the known resolvent cubic root $s=1$ put into the degree 2 and degree 1 equations.  Thus the linear system for the constant coefficients gives $t_1=6$, $t_2=-2$.
