Degrees, zero's, and multiplicity

[cheddacheese]

f(x)= -4(x^2+3)^3

-When I mulitply everything out, I come out with a degree of six. But in my other problems, you have to add them. So if I added the degree would be 5. So I'm not sure which is the correct one.

-as for finding the zeros, I am stuck. Can someone point me in the right direction. When i try to find set (x^2+3)=0 I get a negative in the square root. I'm probably doing this wrong.

 

[BigGlenntheHeavy]

You have a polynomial of degree 6 and it never reaches the x axis.

 

[cheddacheese]

So, it it never reaches the x-axis then that means there are no zeros and i can't put a multiplicity to it?

 

[stapel]

When i try to find set (x^2+3)=0 I get a negative in the square root. I'm probably doing this wrong.

If you aren't familiar with complex numbers (which is how you deal with negatives inside square roots), then you probably don't have to find those zeroes. :wink:

 

[DrMike]

When I mulitply everything out, I come out with a degree of six. But in my other problems, you have to add them. So if I added the degree would be 5. So I'm not sure which is the correct one.

Remember that (x^2+3)^3 means (x^2+3)*(x^2+3)*(x^2+3). When you add the degrees of the polynomial you multiply together, you get 2+2+2 = 6.

as for finding the zeros, I am stuck. Can someone point me in the right direction. When i try to find set (x^2+3)=0 I get a negative in the square root. I'm probably doing this wrong.

Are you familar with complex numbers? The roots are \(\displaystyle {\pm\sqrt{-3}}=\pm{\sqrt 3}i\). But these are complex numbers. If you only want reall roots, there are none.

 

[cheddacheese]

Thank you all so much for you help!!! I really appreciate it! I was just getting myself confused. Thank you again!