Question

solve each inequality.
\\(\frac{(x + 2)(x - 7)(x + 10)}{x - 5} \leq 0\\)

Explanation:

Step1: Find critical points

Set numerator and denominator zero:
Numerator: \((x + 2)(x - 7)(x + 10)=0\) gives \(x=-10, -2, 7\).
Denominator: \(x - 5 = 0\) gives \(x = 5\) (undefined here).

Step2: Divide number line into intervals

Intervals: \((-\infty, -10)\), \((-10, -2)\), \((-2, 5)\), \((5, 7)\), \((7, \infty)\).

Step3: Test sign in each interval

• For \(x < -10\) (e.g., \(x=-11\)): \(\frac{(-)(-)(-)}{(-)}=\frac{(-)}{(-)}=+\), not \(\leq0\).
• For \(-10 < x < -2\) (e.g., \(x=-5\)): \(\frac{(+)(-)(-)}{(-)}=\frac{(+)}{(-)}=- \), satisfies \(\leq0\).
• For \(-2 < x < 5\) (e.g., \(x=0\)): \(\frac{(+)(-)(+)}{(-)}=\frac{(-)}{(-)}=+\), not \(\leq0\).
• For \(5 < x < 7\) (e.g., \(x=6\)): \(\frac{(+)(-)(+)}{(+)}=\frac{(-)}{(+)}=- \), satisfies \(\leq0\).
• For \(x > 7\) (e.g., \(x=8\)): \(\frac{(+)(+)(+)}{(+)}=+\), not \(\leq0\).

Step4: Check critical points

• \(x=-10\): \(\frac{0}{-15}=0\), satisfies \(\leq0\).
• \(x=-2\): \(\frac{0}{-7}=0\), satisfies \(\leq0\).
• \(x=5\): Undefined, exclude.
• \(x=7\): \(\frac{(+)(0)(+)}{(+)}=0\), satisfies \(\leq0\).

Answer:

\( -10 \leq x \leq -2 \); \( 5 < x \leq 7 \)