Solve the inequality (x - 2)^3 (x - 4)^4 (x - 5)^3 (x - 10)^6 <= 0.

Step-by-step solution

1. 1

   List the zeros and their multiplicities. The product vanishes exactly at

   $x=2\ (\text{mult. }3),\quad x=4\ (\text{mult. }4),\quad x=5\ (\text{mult. }3),\quad x=10\ (\text{mult. }6).$

   Because the inequality is $\le$ rather than $<$, every one of these four points is automatically a solution — a detail that turns out to matter a lot at $x=10$.

2. 2

   Use parity of the exponents to shrink the problem. An even power is never negative, so $(x-4)^{4}$ and $(x-10)^{6}$ are $\ge 0$ everywhere and cannot flip the sign anywhere. Away from the zeros, the sign of the whole product is therefore the sign of

   $(x-2)^{3}(x-5)^{3},$

   which in turn has the same sign as $(x-2)(x-5)$ since cubing preserves sign.

3. 3

   Read the sign of $(x-2)(x-5)$. This upward parabola is negative strictly between its roots and positive outside:

   $(x-2)(x-5)<0\iff 2<x<5.$

   So the product is negative on $(2,5)$ — except that at $x=4$ it is zero, which still satisfies $\le 0$, so nothing is lost there — and positive on $(-\infty,2)\cup(5,10)\cup(10,\infty)$.

4. 4

   Assemble the solution set, including the isolated point. Combining the negative interval with all four zeros:

   $[2,5]\cup\{10\}.$

   The point $x=10$ is isolated: on both sides of it the product is strictly positive (it is a sixth power multiplied by positive factors), yet at $x=10$ itself the product equals $0$, which the $\le$ admits. Dropping this point is the single most common error on this type of question.

5. 5

   Verify with sample values. $x=0$: $(-8)(256)(-125)(10^{6})>0$, excluded ✓. $x=3$: $(1)(1)(-8)(\ldots)=-941192<0$, included ✓. $x=4.5$: $\approx-3379<0$, included ✓. $x=6$: $+4194304>0$, excluded ✓. $x=9$: $+13720000>0$, excluded ✓. $x=10$: exactly $0$, included ✓. $x=11$: $+378071064>0$, excluded ✓.