Solve the following inequality:

$$
x^{4}-4 x^{3}+8 x^{2}-8 x \leq 96
$$
I. solution. Transform the above inequality as follows:

$$
x^{4}-4 x^{3}+8 x^{2}-8 x-96 \leq 0
$$

As is well known, if an integer-coefficient polynomial has an integer root, then it is a divisor of its constant term. Thus, it is worth examining the divisors of -96, in the hope of finding a solution among them.

In the first case, $-96=(-1) \cdot 96=-96=1 \cdot(-96)$. By substituting, we can see that none of the divisors are solutions.

In the second case, $-96=(-2) \cdot 48=2 \cdot(-48)$. By substituting, we can see that $x=-2$ is a root of the equation. By polynomial division, we get the following cubic function:

$$
\left(x^{4}-4 x^{3}+8 x^{2}-8 x-96\right):(x+2)=x^{3}-6 x^{2}+20 x-48
$$

Applying the above method further: $-96=4 \cdot(-24)$ and $-96=(-4) \cdot 24$. By substituting, we can see that $x=4$ is also a root of the equation. By another polynomial division based on the already obtained cubic function, we get the following:

$$
\left(x^{3}-6 x^{2}+20 x-48\right):(x-4)=x^{2}-2 x+12
$$

However, for this quadratic polynomial, the discriminant value is $4-4 \cdot 120$ and $f(0)=-96<0$, so the function takes positive values on the $(-\infty ;-2)$ and $(4 ; \infty)$ intervals, and negative values on the $(-2 ; 4)$ interval. Since its graph does not intersect the $x$-axis elsewhere, its value cannot change from positive to negative or vice versa at any other point. Since the inequality in the problem allows for equality as well, it is satisfied when $-2 \leq x \leq 4$.

Remark. Many people graphed the function $f(x)$ using a computer program and read off the two roots. However, it is still necessary to verify by substitution that the two read-off values are indeed roots (since the actual root could be, for example, $-2.00004$), and to ensure that there are no other solutions (since no matter what program we use to graph it, the curve of the function is only visible in a certain interval).

Pszota Máté (Révkomárom, Selye János Gimn., 12th grade)

II. solution. With the same transformations, the left side of the inequality can be factored as follows:

$$
\left(x^{2}-2 x\right)\left(\left(x^{2}-2 x\right)+4\right)
$$

Let's introduce a new variable: $y=x^{2}-2 x=x(x-2)$, with this the inequality to be solved takes the following simple form:

$$
y^{2}+4 y \leq 96
$$

Add 4 to both sides of the inequality, thus obtaining a perfect square:

$$
(y+2)^{2} \leq 100
$$

The solution to this is: $-10 \leq y+2 \leq 10$, from which $0 \leq y \leq 8$ or $-12 \leq y<0$.

Let's return to the original variable. We get the following inequalities:

$$
-12 \leq x(x-2)<0, \quad \text { and } \quad 0 \leq x(x-2) \leq 8
$$

The function $f(x)=x(x-2)$ takes negative and zero values on the interval $[0 ; 2]$, its minimum value is $f\left(\frac{0+2}{2}\right)=f(1)=$ -1, so it is true everywhere that $x(x-2) \geq-12$. Therefore, the first inequality $x(x-2)<0$ is satisfied when $x \in(0 ; 2)$.

The function being examined takes positive values outside the above interval, i.e., when $x \notin(0,2)$, in which case the inequality $x(x-2) \leq 8$ must be satisfied. The roots of the equation $x^{2}-2 x-8=0$ are

$$
\frac{2-\sqrt{4+4 \cdot 8}}{2}=-2 \quad \text { and } \quad \frac{2+\sqrt{4+4 \cdot 8}}{2}=4
$$

so if $x \notin(0,2)$, then the solution to the inequality $x(x-2) \leq 8$ is $x \in[-2 ; 0] \cup[2 ; 4]$.

Finally, the solution to the original inequality is $x \in[-2 ; 4]$.

Werner András (Budapest, Piarista Gimn., 10th grade)

Remark. Many solved the problem on our website in the following way: by transforming the inequality into

$$
(x-1)^{4}+2(x-1)^{2}-99 \leq 0
$$

form, and then introducing the variable $y=(x-1)^{2}$.