Question

solve the following inequality algebraically.
$x^3 - 6x^2 > 0$
the solution is
(type your answer in interval notation. simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Factor the inequality

Factor \(x^3 - 6x^2\) to get \(x^2(x - 6)>0\).

Step2: Analyze the factors

The factor \(x^2\) is always non - negative (since any real number squared is non - negative). \(x^2 = 0\) when \(x = 0\). For the product \(x^2(x - 6)\) to be greater than 0, we need to consider when both factors satisfy the inequality conditions. Since \(x^2\geq0\) for all real \(x\), and we want \(x^2(x - 6)>0\), we note that \(x^2>0\) when \(x
eq0\) and we also need \(x - 6>0\) (because if \(x^2>0\) and \(x - 6>0\), their product will be positive; if \(x^2 = 0\), the product is 0 which does not satisfy the inequality \(>0\)).
Solving \(x - 6>0\) gives \(x>6\). We also need to check the case when \(x^2 = 0\) (i.e., \(x = 0\)): when \(x = 0\), the left - hand side of the inequality \(x^3-6x^2=0 - 0 = 0\), which does not satisfy \(x^3 - 6x^2>0\). When \(x<6\) and \(x
eq0\), if \(x\) is between 0 and 6 (e.g., \(x = 1\)), then \(x^2>0\) and \(x - 6<0\), so \(x^2(x - 6)<0\); if \(x<0\) (e.g., \(x=-1\)), \(x^2>0\) and \(x - 6<0\), so \(x^2(x - 6)<0\).

Answer:

\((6,\infty)\)