Solve the chained inequality (x + 1)(x + 2) >= (x + 3)(x + 4) >= (x + 5)(x + 6) and write the solution set as an interval.

Step-by-step solution

1. 1

   Split the chain into two separate inequalities. A statement of the form $A\ge B\ge C$ is shorthand for two conditions that must hold at the same time:

   $\text{(I)}\quad (x+1)(x+2)\ge(x+3)(x+4),\qquad \text{(II)}\quad (x+3)(x+4)\ge(x+5)(x+6)$

   The final answer is the intersection of the two solution sets, not their union. It is tempting to jump straight to comparing the outer expressions $(x+1)(x+2)$ and $(x+5)(x+6)$, but that would be a weaker condition and would produce a wrong (too large) answer set.

2. 2

   Expand both sides of (I) and watch the quadratic terms cancel. Multiplying out,

   $x^2+3x+2\ \ge\ x^2+7x+12$

   Both sides have exactly the same leading term $x^2$, so subtracting $x^2$ from both sides is legal and leaves a linear inequality:

   $3x+2\ \ge\ 7x+12$

   This cancellation is the whole point of the problem: because consecutive shifted products all have leading coefficient $1$, no sign chart or parabola analysis is needed.

3. 3

   Solve the linear inequality (I). Collect the $x$'s on one side:

   $3x+2\ge 7x+12\ \Longrightarrow\ -10\ge 4x\ \Longrightarrow\ x\le-\frac{10}{4}=-\frac{5}{2}$

   Dividing by the positive number $4$ does not flip the inequality sign; the direction only flips when you divide or multiply by a negative number. Reading $-\tfrac52\ge x$ from right to left gives $x\le-\tfrac52$.

4. 4

   Do the same with (II). Expanding,

   $x^2+7x+12\ \ge\ x^2+11x+30\ \Longrightarrow\ 7x+12\ \ge\ 11x+30$

   $-18\ \ge\ 4x\ \Longrightarrow\ x\le-\frac{18}{4}=-\frac{9}{2}$

5. 5

   Intersect the two conditions. We need $x\le-\tfrac52$ and $x\le-\tfrac92$ simultaneously. Since $-\tfrac92=-4.5$ lies to the left of $-\tfrac52=-2.5$, the second condition is the stricter one and it automatically implies the first:

   $x\le-\frac{9}{2}\quad\Longleftrightarrow\quad x\in\left(-\infty,\,-\frac{9}{2}\right]$

6. 6

   Test one value inside and one value outside. At $x=-5$: $(-4)(-3)=12$, $(-2)(-1)=2$, $(0)(1)=0$, and $12\ge2\ge0$ holds. ✓ At $x=-4$ (which satisfies (I) but not (II)): $(-3)(-2)=6$, $(-1)(0)=0$, $(1)(2)=2$, and $0\ge2$ is false. ✗ This confirms that intersecting rather than unioning was the right move, and that the endpoint $x=-\tfrac92$ is included because both inequalities allow equality.