Solve the inequality (x + 1)(x + 3) >= (x - 2)(x - 4).

Step-by-step solution

1. 1

   Expand before deciding what kind of inequality this is. A product on each side suggests a quadratic inequality needing a sign chart, but you cannot know until the two sides are expanded. Do not be tempted to compare factors directly — $(x+1)\ge(x-2)$ and $(x+3)\ge(x-4)$ would be a different (and invalid) argument.

2. 2

   Expand each side with FOIL.

   $(x+1)(x+3)=x^2+4x+3$

   $(x-2)(x-4)=x^2-6x+8$

   so the inequality reads

   $x^2+4x+3\ge x^2-6x+8$

3. 3

   Subtract $x^2$ from both sides — this is the key simplification. Both sides have exactly the same leading term, so it cancels and the inequality becomes linear:

   $4x+3\ge -6x+8$

   No sign chart, no critical points, no parabola: the quadratic character was an illusion.

4. 4

   Collect and isolate $x$. Add $6x$, subtract $3$, divide by the positive $10$:

   $10x+3\ge 8\ \Longrightarrow\ 10x\ge 5\ \Longrightarrow\ x\ge\frac12$

   Because $10>0$ the direction of $\ge$ is preserved, and the endpoint is included since the original inequality was non-strict.

5. 5

   State the answer and check the boundary. The solution set is $\left[\tfrac12,\infty\right)$. At $x=\tfrac12$: left $=\tfrac32\cdot\tfrac72=\tfrac{21}{4}$ and right $=\left(-\tfrac32\right)\left(-\tfrac72\right)=\tfrac{21}{4}$ — equal, so $\tfrac12$ belongs to the set. At $x=0$: left $=3$, right $=8$, and $3\ge 8$ is false, confirming values below $\tfrac12$ fail.