Solve the inequality (2 - 3x)(2x + 1) >= 0.

Step-by-step solution

1. 1

   Keep it factored — do not expand. The expression is already a product of two linear factors, and a product's sign is determined entirely by the signs of its factors. Expanding to $-6x^2+x+2$ would only make the work harder.

2. 2

   Find where each factor is zero.

   $2-3x=0\Rightarrow x=\frac23,\qquad 2x+1=0\Rightarrow x=-\frac12$

   These critical points split the line into $\left(-\infty,-\tfrac12\right)$, $\left(-\tfrac12,\tfrac23\right)$, and $\left(\tfrac23,\infty\right)$.

3. 3

   Test one point in each interval. At $x=-1$: $(2+3)(-2+1)=(5)(-1)=-5<0$. At $x=0$: $(2)(1)=2>0$. At $x=1$: $(-1)(3)=-3<0$. So the pattern is negative, positive, negative.

4. 4

   Explain the pattern from the leading coefficient. Expanding would give a leading term $-6x^2$, i.e. a downward parabola. A downward parabola is positive between its roots and negative outside — the mirror image of the usual case. Reflexively writing "outside the roots" here is the classic mistake.

5. 5

   Include the endpoints because the inequality is non-strict. At $x=-\tfrac12$ and $x=\tfrac23$ the product is exactly $0$, which satisfies $\ge0$. So the solution is the closed interval

   $-\frac12\le x\le\frac23$

6. 6

   Verify by scanning. Evaluating the product at $6001$ points from $-3$ to $3$ and comparing with the claimed interval agrees at every point ✓, and both endpoints return exactly $0$ ✓.