Question

25 multiple choice 1 point 25) solve ( x^3 - 9x^2 - 36x > 0 ). a. ( (-3, 12) ) b. ( (-infty, -3) cup (0, 12) ) c. ( (-infty, -3) cup (12, infty) ) d. ( (-3, 0) cup (12, infty) )

Explanation:

Step1: Factor out common term

$x^3 - 9x^2 - 36x = x(x^2 - 9x - 36)$

Step2: Factor quadratic expression

$x^2 - 9x - 36 = (x + 3)(x - 12)$, so the full factorization is $x(x + 3)(x - 12) > 0$

Step3: Find critical points

Set each factor to 0: $x=0$, $x=-3$, $x=12$

Step4: Test intervals

• For $(-\infty, -3)$: $(-)(-)(-)=(-) < 0$
• For $(-3, 0)$: $(-)(+)(-)=+ > 0$
• For $(0, 12)$: $(+)(+)(-)=(-) < 0$
• For $(12, \infty)$: $(+)(+)(+)=+ > 0$

Answer:

D. $(-3,0) \cup (12, \infty)$