# Thread: How do you find the zeros?

1. ## How do you find the zeros?

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!

2. Just by inspection you can see that 0 is a root. To see if there are any other roots, put $\displaystyle y = x^2$. Then your equation turns into $\displaystyle -x(x^4 - 5x^2 + 4) = -x(y^2 - 5y + 4) = 0$. Can you factor the expression that is quadratic in y?

3. Originally Posted by s3a
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!
You can factor further

$\displaystyle (x^4-5x^2+4)=(x^2-4)(x^2-1)=(x-2)(x+2)(x-1)(x+1)$

So the result:

$\displaystyle -x(x-2)(x+2)(x-1)(x+1)=0\Rightarrow{x}=0,\pm2,\pm1$

4. Originally Posted by JG89
Just by inspection you can see that 0 is a root. To see if there are any other roots, put $\displaystyle y = x^2$. Then your equation turns into $\displaystyle -x(x^4 - 5x^2 + 4) = -x(y^2 - 5y + 4) = 0$. Can you factor the expression that is quadratic in y?
Ok so from -x(y^2 - 5y + 4) = 0 I got: x=0, y=1, y=4. Is this correct? If so, what do I do from here?

5. Originally Posted by s3a
Ok so from -x(y^2 - 5y + 4) = 0 I got: x=0, y=1, y=4. Is this correct? If so, what do I do from here?
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

$\displaystyle (x^2)^2-5x^2+4$

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.

6. Originally Posted by s3a
Ok so from -x(y^2 - 5y + 4) = 0 I got: x=0, y=1, y=4. Is this correct? If so, what do I do from here?

remember , you let x^2=y

so now $\displaystyle x^2=1\Rightarrow$ x=1 , -1

$\displaystyle x^2=4\Rightarrow$ x=2 , -2

Now you can proceed to find its zeros .

7. Originally Posted by VonNemo19
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

$\displaystyle (x^2)^2-5x^2+4$

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.

8. Originally Posted by s3a
Edit: I just saw the factoring further post. I think that's what I needed. I'll try to complete the problem now.
Great! I'm glad you see it!

Look at my first post. You should be able to decipher it now. If not, let me know.

9. Originally Posted by VonNemo19
Great! I'm glad you see it!

Look at my first post. You should be able to decipher it now. If not, let me know.
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.