Question

solve the following inequality. (x - 3)^2(x + 9) < 0 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: Analyze $(x - 3)^2$ property

The square of any real - number is non - negative. That is, $(x - 3)^2\geq0$ for all real $x$, and $(x - 3)^2 = 0$ when $x = 3$.

Step2: Analyze the inequality condition

Since $(x - 3)^2\geq0$, for the product $(x - 3)^2(x + 9)<0$ to hold, we need $(x - 3)^2
eq0$ (because if $(x - 3)^2=0$, the product is 0) and $x + 9<0$.
If $(x - 3)^2
eq0$, then $x
eq3$. And from $x + 9<0$, we solve for $x$:
$x+9<0$ implies $x<-9$.

Answer:

$(-\infty,-9)$