How do you find the zeros?

• August 27th 2009, 07:54 AM
s3a
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!
• August 27th 2009, 07:57 AM
JG89
Just by inspection you can see that 0 is a root. To see if there are any other roots, put $y = x^2$. Then your equation turns into $-x(x^4 - 5x^2 + 4) = -x(y^2 - 5y + 4) = 0$. Can you factor the expression that is quadratic in y?
• August 27th 2009, 08:08 AM
VonNemo19
Quote:

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

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

So the result:

$-x(x-2)(x+2)(x-1)(x+1)=0\Rightarrow{x}=0,\pm2,\pm1$
• August 27th 2009, 08:14 AM
s3a
Quote:

Originally Posted by JG89
Just by inspection you can see that 0 is a root. To see if there are any other roots, put $y = x^2$. Then your equation turns into $-x(x^4 - 5x^2 + 4) = -x(y^2 - 5y + 4) = 0$. 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?
• August 27th 2009, 08:16 AM
VonNemo19
Quote:

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

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

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.
• August 27th 2009, 08:30 AM
Quote:

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 $x^2=1\Rightarrow$ x=1 , -1

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

Now you can proceed to find its zeros .
• August 27th 2009, 08:35 AM
s3a
Quote:

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

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

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.
• August 27th 2009, 08:40 AM
VonNemo19
Quote:

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.(Hi)
• August 27th 2009, 09:01 AM
s3a
Quote:

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.(Hi)

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.