Hello, Mark!
Find the set of values so that $\displaystyle P(x) \:=\: 3x^3 + 29x^2 + 65x  25 \:>\:0$
given that $\displaystyle (x+5)$ is a repeated factor and $\displaystyle (3x  1)$ is the third factor. They factored the polynomial for us: .$\displaystyle P(x) \;=\;(x+5)^2(3x1)$
When is the polynomial greater than zero?
It is equal to zero when: .$\displaystyle (x+5)^2(3x1) \:=\:0$
. . which has roots: .$\displaystyle x \:=\:\text{}5,\:\frac{1}{3}$
These two values divide the number line into three intervals:
. . $\displaystyle \underbrace{\;   \; }\;(\text{}5)\;\underbrace{\;   \;}\;\left(\tfrac{1}{3}\right)\;\underbrace{\;   \; }$
Test a value in each interval:
. . $\displaystyle \begin{array}{ccccc}P(\text{}6) &=& (\text{}1)^2(\text{}19) &=& \text{neg} \\ \\[4mm]
P(0) &=& (5)^2(\text{}1) &=& \text{neg} \\ \\[4mm]
P(1) &=& (6^2)(1) &=& \text{pos} \end{array}$
Therefore, $\displaystyle P(x)$ is positive for: .$\displaystyle x \:>\:\tfrac{1}{3}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If we graph the polynomial equation, the answer is obvious . . . Code:

 *


 *
5 
     o       +   o  
* *  * 1/3
* *  *
* * *
* 
* 
