Solve the inequality (2x - 1)(x + 1)/(x^2 + x + 2) <= 0.

Step-by-step solution

1. 1

   Examine the denominator first. For $x^{2}+x+2$ the discriminant is

   $\Delta=1^{2}-4(1)(2)=1-8=-7<0,$

   and the leading coefficient is positive, so the parabola never crosses the axis:

   $x^{2}+x+2>0\quad\text{for every real }x.$

   That single fact does most of the work — there are no excluded points and no sign changes from below the bar.

2. 2

   Reduce to a polynomial inequality. Multiplying both sides by a strictly positive quantity preserves the inequality, so

   $\frac{(2x-1)(x+1)}{x^{2}+x+2}\le0\iff(2x-1)(x+1)\le0.$

   This equivalence would fail if the denominator could change sign, which is exactly why step 1 had to come first.

3. 3

   Find the critical points. Setting each factor to zero:

   $2x-1=0\Rightarrow x=\tfrac12,\qquad x+1=0\Rightarrow x=-1.$

   Both make the numerator zero and the denominator nonzero, so the whole fraction is $0$ there — which satisfies $\le0$ and means both endpoints are included.

4. 4

   Read the sign of the quadratic. The product $(2x-1)(x+1)$ is an upward-opening parabola, so it is negative between its roots and positive outside. Test points confirm it: at $x=-2$, $(-5)(-1)=5>0$; at $x=0$, $(-1)(1)=-1<0$; at $x=1$, $(1)(2)=2>0$.

5. 5

   State the solution set. Combining the negative interval with the two zeros:

   $x\in\left[-1,\ \tfrac12\right].$

   Spot checks on the original fraction: at $x=-2$ it is $\tfrac{5}{4}>0$ (excluded) ✓, at $x=0$ it is $-\tfrac12<0$ (included) ✓, at $x=\tfrac12$ it is $0$ (included) ✓, and at $x=1$ it is $\tfrac12>0$ (excluded) ✓.