<s>[INST] 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?
 [/INST] @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 $b'$.
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