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!
Substitute back for x and solve. But I must say that when looking to factor polynomials like these, I find it easier to recognize the fact that it is "quadratic". Because
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.
Ok now I see that
-x((sqrt(1))^4 - 5(2)^2 + 4) = 0
and
-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.
I was told the answer for #10 2) was C. (Problem I'm currently doing). I got it mostly right but I'm wondering why I have a portion wrong. I don't know what that table method is called but I tried to explain it in case it's not something universal. The portions I'm talking about are the "WTF" portions lol.