1. If , then and .
If then , . Both of these solutions satisfy the original equation.
2. Factor to . Exactly one or three of these factors must be positive. Use casework.
Sort of, but you're not solving for x. If x is negative, then x-3, x-5, and x-6 are negative, and the product is positive. However we are only interested in the intervals where LHS is negative. Hence is not part of the solution set.
Repeat the same for , , , and .
I thought I explained this...
There are five important intervals you need to test, namely all the intervals around the roots, so .
So choose a value of x in the first interval, say x = -7. Then we have
Since this value is positive, then we can say that the region in which x < -6 does not satisfy .
Now try testing values of x from each of the other important regions. Accept the regions which give negative values for the function.
The equation is: .
Go here: x(x-3)(x-5)(x-6) - Wolfram|Alpha to look at your polynomial.
Now found the values of x that make the graph go below the x-axis - as that's essentially what the equation is saying.
So looking at the graph, it's obvious that values of is negative when
Do you see how I got that? Now you find the next solution