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?
.In this question, you have to attempt to complete two squares, namely:
$$
x^4-4x^3 + 12x^2-24x+24 = x^4 -4x^3+6x^2-4x+1 + 6x^2-20x+23 
\\  = (x-1)^4 + 6x^2 -20x+23
$$

Now, as it turns out, we can complete the second square, and: 
$$
6x^2-20x+23 = \frac{2}{3}(3x-5)^2 + \frac{19}{3} 
$$

Hence, we can rewrite the whole expression as:
$$
x^4-4x^3 + 12x^2-24x+24 = (x-1)^4 + \frac{2}{3}(3x-5)^2 + \frac{19}{3}
$$

It is a sum of positive expressions, hence is always positive, hence $P(x) = |P(x)|$.
The first step is definitely recalling a modulus function's definition:
$$|x| := \begin{cases}x &amp; \text{if } x\ge 0 \\ -x &amp; \text{if } x\le 0\end{cases}$$
So the requirement $P(x) = |P(x)|$ is equivalent to $P(x) \ge 0$.

Hope you can continue from that.