Question

solve the inequality algebraically.
$2x^3 > -16x^2$
move all terms to the left side of the inequality and then define the left side of the inequality as f(x). then list the intervals determined by the real zeros of f. (type your answers in interval notation. use ascending order.)
interval
sign
what is the solution?
$2x^3 > -16x^2$ is $(-8,0)\cup(0,\infty)$
(type your answer in interval notation. simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Move all terms to left

Start with \(2x^3 > -16x^2\). Add \(16x^2\) to both sides: \(2x^3 + 16x^2 > 0\). Factor out \(2x^2\): \(2x^2(x + 8) > 0\).

Step2: Find critical points

Set each factor equal to zero: \(2x^2 = 0\) gives \(x = 0\) (double root), and \(x + 8 = 0\) gives \(x = -8\). Critical points are \(x = -8\) and \(x = 0\).

Step3: Test intervals

Intervals: \((-\infty, -8)\), \((-8, 0)\), \((0, \infty)\).

• For \(x < -8\) (e.g., \(x = -9\)): \(2(-9)^2(-9 + 8)=2(81)(-1)= -162 < 0\).
• For \(-8 < x < 0\) (e.g., \(x = -1\)): \(2(-1)^2(-1 + 8)=2(1)(7)= 14 > 0\).
• For \(x > 0\) (e.g., \(x = 1\)): \(2(1)^2(1 + 8)=2(1)(9)= 18 > 0\).

Note: \(x = 0\) makes \(2x^2(x + 8) = 0\), so it's not included.

Step4: Determine solution

The inequality \(2x^2(x + 8) > 0\) holds when \(-8 < x < 0\) or \(x > 0\). In interval notation: \((-8, 0) \cup (0, \infty)\).

Answer:

\((-8, 0) \cup (0, \infty)\)