Solve the inequality $\frac{(x-1)^2(x+1)^3}{x^4(x-2)}\le 0$

Expert intro: This is a rational inequality, so the key idea is to find where the expression is zero or undefined, then use sign analysis on the intervals between critical points.

Detailed walkthrough

Step 1: Find the critical points

Consider

$\frac{(x-1)^2(x+1)^3}{x^4(x-2)}\le 0.$

The numerator is zero at:

• $x=-1$ from $(x+1)^3$
• $x=1$ from $(x-1)^2$

The denominator is zero at:

• $x=0$ from $x^4$
• $x=2$ from $(x-2)$

So the critical points are $-1,0,1,2$.

Step 2: Determine sign changes

• $(x-1)^2$ has even power, so it does not change sign at $x=1$.
• $x^4$ has even power, so it does not change sign at $x=0$.
• $(x+1)^3$ has odd power, so it does change sign at $x=-1$.
• $(x-2)$ has odd power, so it does change sign at $x=2$.

Step 3: Test intervals

The sign of the expression is:

• positive on $(-\infty,-1)$
• negative on $(-1,0)$
• undefined at $x=0$
• negative on $(0,2)$ except that the sign does not change at $x=1$
• positive on $(2,\infty)$

Since we need $\le 0$, include the intervals where the expression is negative and also where it equals zero.

Step 4: Include zeros of the numerator

• At $x=-1$, the expression equals $0$, so include it.
• At $x=1$, the expression equals $0$, so include it.
• At $x=0$ and $x=2$, the expression is undefined, so exclude them.

Final answer

$[-1,0)\cup(0,2)$

💡 Pitfall guide

Do not include points where the denominator is zero, even if the sign chart looks favorable. Also remember that even powers like $(x-1)^2$ and $x^4$ do not change the sign of the expression across their zeros.

🔄 Real-world variant

If the inequality were $<0$ instead of $\le 0$, the solution would exclude the zeros of the numerator, so the endpoints $x=-1$ and $x=1$ would not be included. If the denominator had $(x-2)^2$ instead of $(x-2)$, the sign change at $x=2$ would disappear.

🔍 Related terms

rational inequality, sign chart, critical points