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

what is the solution?
(type your answer in interval notation. simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Rearrange the inequality

Start with the inequality \(2x^{3} > -16x^{2}\). Move all terms to the left - hand side:
\(2x^{3}+ 16x^{2}>0\)
Factor out the greatest common factor, which is \(2x^{2}\):
\(2x^{2}(x + 8)>0\)

Step2: Find the critical points

To find the critical points, set each factor equal to zero:

• For \(2x^{2}=0\), we get \(x = 0\) (with multiplicity 2).
• For \(x + 8=0\), we get \(x=-8\).

The critical points divide the real number line into the intervals \((-\infty,-8)\), \((-8,0)\), and \((0,\infty)\).

Step3: Test the intervals

• Interval \((-\infty,-8)\): Let's choose a test point, say \(x=-9\).

Substitute into \(2x^{2}(x + 8)\): \(2\times(-9)^{2}\times(-9 + 8)=2\times81\times(-1)=-162<0\)

• Interval \((-8,0)\): Let's choose a test point, say \(x = - 1\).

Substitute into \(2x^{2}(x + 8)\): \(2\times(-1)^{2}\times(-1 + 8)=2\times1\times7 = 14>0\)

• Interval \((0,\infty)\): Let's choose a test point, say \(x = 1\).

Substitute into \(2x^{2}(x + 8)\): \(2\times(1)^{2}\times(1 + 8)=2\times1\times9=18>0\)

But we also need to consider the original inequality \(2x^{3}+16x^{2}>0\). Note that when \(x = 0\), the left - hand side is \(2\times0^{3}+16\times0^{2}=0\), which does not satisfy the inequality \(>0\). When \(x=-8\), the left - hand side is \(2\times(-8)^{3}+16\times(-8)^{2}=2\times(-512)+16\times64=-1024 + 1024 = 0\), which also does not satisfy the inequality \(>0\).

Answer:

The solution to the inequality \(2x^{3}>-16x^{2}\) is \((-8,0)\cup(0,\infty)\)