<s>[INST] Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$ $$P(x)=x^{4}-4x^{3}+12x^{2}-24x+24$$
I'd like to prove that $P(x)=|P(x)|$. I don't know where to begin. What would be the first step?
 [/INST] By completing the square for $x^4+12x^2\cdots$,
$$P(x)=(x^2+6)^2-4x^3-24x-12=(x^2+6)^2-4x(x^2+6)-12\\
=(x^2+6)(x^2-4x+6)-12.$$
The minimum value of the first factor is $6$ and that of the second is $2$.

Or completing the square for $x^2(x^2-4x\cdots)$,
$$P(x)=x^2(x-2)^2+8x^2-24x+24=x^2(x-2)^2+8(x^2-3x+3).$$
As you can easily check, the trinomial on the right has no real root.

Or completing the fourth power for $x^4-4x^3\cdots$,
$$P(x)=(x-1)^4+P(x)-(x^4-4x^3+6x^2-4x+1)\\=(x-1)^4+6x^2-20x+23$$ and the trinomial on the right has no real root.
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