Solve the inequality (2x + 1)(3x - 1)^2 divided by (x^3 times (root 2 - 1)) is less than or equal to 0.

Step-by-step solution

1. 1

   Deal with the constant factor first. $\sqrt2\approx 1.4142>1$, so $\sqrt2-1>0$. Dividing by a positive constant does not change the direction of an inequality, so it can simply be dropped:

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

2. 2

   Classify each factor by how it affects the sign. The critical values are $x=-\tfrac12$, $x=0$ and $x=\tfrac13$.

   • $(2x+1)$ changes sign at $-\tfrac12$;
   • $(3x-1)^{2}$ has an even power, so it is $\ge 0$ everywhere and never flips the sign — it only vanishes at $\tfrac13$;
   • $x^{3}$ has an odd power, so it flips the sign at $0$, exactly like $x$ does;
   • $x=0$ is excluded from the domain.
3. 3

   Reduce to the sign of a simpler quotient. Away from $x=\tfrac13$ the factor $(3x-1)^2$ is strictly positive, so on the rest of the line the inequality has the same sign behaviour as

   $\frac{2x+1}{x}\le 0.$

4. 4

   Build the sign chart on the intervals cut by -1/2 and 0.

   $x<-\tfrac12:\ \tfrac{(-)}{(-)}=+\quad;\quad -\tfrac12<x<0:\ \tfrac{(+)}{(-)}=-\quad;\quad x>0:\ \tfrac{(+)}{(+)}=+.$

   So the quotient is negative exactly on $\left(-\tfrac12,0\right)$.

5. 5

   Add the equality cases. The inequality is non-strict, so include every point where the expression equals zero and is defined: $x=-\tfrac12$ (from $2x+1=0$) and $x=\tfrac13$ (from $(3x-1)^2=0$). The point $x=0$ stays excluded because the denominator vanishes there.

6. 6

   Write the solution set.

   $x\in\left[-\tfrac12,\,0\right)\cup\left\{\tfrac13\right\}.$

   The isolated point $\tfrac13$ is the feature worth remembering: an even-power factor contributes no sign change, but in a non-strict inequality it still contributes its own root as a lone solution. Spot-checks confirm it — at $x=\tfrac13$ the expression is $0$, and at $x=0.4$ it is positive.