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.
complete the following table.
(type your answers in interval notation. use ascending order.)

interval

| sign | $\boldsymbol{
abla}$ | $\boldsymbol{
abla}$ | $\boldsymbol{
abla}$ |
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 to zero: \( 2x^2 = 0 \) gives \( x = 0 \), and \( x + 8 = 0 \) gives \( x = -8 \). These divide the number line into intervals: \( (-\infty, -8) \), \( (-8, 0) \), and \( (0, \infty) \).

Step3: Test intervals

• For \( (-\infty, -8) \), pick \( x = -9 \): \( 2(-9)^2(-9 + 8) = 2(81)(-1) = -162 < 0 \)? No, wait, original inequality after factoring is \( 2x^2(x + 8) > 0 \). \( 2x^2 \) is always non - negative (positive except at \( x = 0 \)). So when \( x < -8 \), \( x + 8 < 0 \), so \( 2x^2(x + 8) < 0 \) (since positive times negative is negative). When \( -8 < x < 0 \), \( x + 8 > 0 \), \( 2x^2>0 \), so \( 2x^2(x + 8)>0 \). When \( x > 0 \), \( x + 8 > 0 \), \( 2x^2>0 \), so \( 2x^2(x + 8)>0 \). At \( x = -8 \), \( 2(-8)^2(-8 + 8)=0 \), not greater than 0. At \( x = 0 \), \( 2(0)^2(0 + 8)=0 \), not greater than 0.

Step4: Determine solution

The inequality \( 2x^2(x + 8)>0 \) holds when \( -8 < x < 0 \) or \( x > 0 \). In interval notation, this is \( (-8, 0)\cup(0, \infty) \). Wait, the initial wrong test was corrected. The key is that \( 2x^2 \) is positive for \( x
eq0 \), so we need \( x + 8>0 \) (since \( 2x^2 \) is positive when \( x
eq0 \)). So \( x + 8>0 \) gives \( x > - 8 \), but we exclude \( x = 0 \) (since at \( x = 0 \), the left - hand side is 0). So the solution is \( (-8, 0)\cup(0, \infty) \).

Answer:

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