27. $(x-1)(3-x)(x-2)^2>0$

Key takeaway: For a product inequality, factor behavior matters: odd powers can change sign, while even powers do not. Here, $(x-2)^2$ stays nonnegative, so the sign comes from the other factors.

Step 1: Identify the critical points

Set each factor equal to zero:

• $x-1=0 \Rightarrow x=1$
• $3-x=0 \Rightarrow x=3$
• $(x-2)^2=0 \Rightarrow x=2$

So the critical points are $1,2,3$.

Step 2: Analyze the sign of each factor

• $(x-2)^2\ge 0$ for all $x$, and it is positive except at $x=2$.
• The sign of the product therefore depends on $(x-1)(3-x)$ away from $x=2$.

Check the intervals:

• If $x<1$, then $(x-1)<0$ and $(3-x)>0$, so the product is negative.
• If $1<x<3$, then $(x-1)>0$ and $(3-x)>0$ when $x<3$? Wait: for $1<x<3$, $(3-x)>0$, so the product is positive.
• If $x>3$, then $(x-1)>0$ and $(3-x)<0$, so the product is negative.

Now exclude $x=2$, where the product equals $0$.

Final answer

$(1,2)\cup(2,3)$

---

Pitfalls the pros know 👇 A common mistake is forgetting that $(x-2)^2$ is never negative, so it does not create a sign flip. Another mistake is including $x=2$ in the answer even though the inequality is strict and the product is zero there.

What if the problem changes? If the inequality were $\ge 0$, then the solution would include the zeros of the expression: $x=1,2,3$ as appropriate, with the sign intervals adjusted to include the positive regions and the zeros.

Tags: product inequality, even power, sign analysis