Question

solve:
\frac{(x + 5)(x + 1)}{x - 4} \geq 0
? \leq x \leq \quad or x > \quad

Explanation:

Step1: Find critical points

Set numerator and denominator to zero:

• Numerator: \((x + 5)(x + 1)=0\) gives \(x=-5\), \(x=-1\).
• Denominator: \(x - 4 = 0\) gives \(x = 4\) (undefined here).

Critical points: \(x=-5\), \(x=-1\), \(x = 4\). These divide the number line into intervals: \((-\infty, -5)\), \((-5, -1)\), \((-1, 4)\), \((4, \infty)\).

Step2: Test intervals

• Interval \((-\infty, -5)\): Pick \(x=-6\).

\(\frac{(-6 + 5)(-6 + 1)}{-6 - 4}=\frac{(-1)(-5)}{-10}=\frac{5}{-10}<0\). Not valid.

• Interval \((-5, -1)\): Pick \(x=-3\).

\(\frac{(-3 + 5)(-3 + 1)}{-3 - 4}=\frac{(2)(-2)}{-7}=\frac{-4}{-7}>0\). Valid.

• Interval \((-1, 4)\): Pick \(x=0\).

\(\frac{(0 + 5)(0 + 1)}{0 - 4}=\frac{5(1)}{-4}<0\). Not valid.

• Interval \((4, \infty)\): Pick \(x=5\).

\(\frac{(5 + 5)(5 + 1)}{5 - 4}=\frac{10(6)}{1}=60>0\). Valid.

Step3: Check critical points

• \(x=-5\): \(\frac{(-5 + 5)(-5 + 1)}{-5 - 4}=\frac{0(-4)}{-9}=0\). Satisfies \(\geq0\), so include \(x=-5\).

• \(x=-1\): \(\frac{(-1 + 5)(-1 + 1)}{-1 - 4}=\frac{4(0)}{-5}=0\). Satisfies \(\geq0\), so include \(x=-1\).

• \(x=4\): Denominator is 0, undefined. Exclude \(x=4\).

Answer:

\(-5 \leq x \leq -1\) or \(x > 4\)