Just by inspection you can see that 0 is a root. To see if there are any other roots, put . Then your equation turns into . Can you factor the expression that is quadratic in y?
I need to find the zeros as well as see where it is positive or negative so that I can match it to a graph for a question. My only problem (from what I can tell so far) is how to find the zeros. When I find the zeros, I think I might be able to manage the rest:
-x^5 + 5x^3 - 4x
-x(x^4 - 5x^2 + 4) <--------------What do I do next?
Any help would be greatly appreciated!
Thanks in advance!
Substitution is a very good way to do it too, but I tend to get confused when another variable is introduced. But, it all boils down to whichever method that you are most comfortable with.
-x((sqrt(1))^4 - 5(2)^2 + 4) = 0
-x((2^4 - 5(2)^2 + 4) = 0
are both true.
So, what does this mean? Since y=1 and y=4, my zeros are x=1 and x=2?
Edit: I just saw the factoring further post. I think that's what I needed. I'll try to complete the problem now.