Finding the Zeros of a Polynomial Function

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To find the zeros of the polynomial function f(x) = x^4 - 25, the equation can be factored as (x^2 - 5)(x^2 + 5) = 0. This leads to two real solutions: x = ±√5, and two complex solutions: x = ±i√5. The initial confusion stemmed from misidentifying -25 as a zero, which is incorrect. The discussion also included a brief unrelated exchange about names and locations. Understanding the algebraic approach is crucial for accurately identifying all zeros of the polynomial.
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hey.. how would i find the zeros of the following function:

f(x)=x^4-25

I tried inputting the value into my caculator and then go to table find y values that equal 0 in x and i only found -25.. how do i find the rest?


thanks
 
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-25 shouldn't be one of the zeros. You can find the two real zeros graphically, but why not do it algebraically?

x4 - 25 = 0
x4 = 25
x2 = {-5, +5}
and so on...
 
x^4-25=0
x^4=25
x^2=5
x=+-/sqrt{5}
 
Last edited:
hey vitaly, do you by any chance live in Ohio? just guessing...
 
Nope... What makes you say that?
 
i knew a person by the name Vitaly there, but i guess it is a common name.. so i was being unrealistic...
 
Oh, well I live in TN...

Is the person you knew Russian? I have never met anybody else with my name before. Most people think it's weird; it's definitely not common here.
 
yup, they were russians.. infact there was two of them cause they were twins...
 
ok that was a stupid question... didnt think of that for some odd reason.. thanks much for ur help1
 
  • #10
For the sake of precision, from:

x^4 = 25

you should deduce

x^2 = 5 or x^2 = -5

and not just the first one.


Actually, I'm a big fan of using factoring instead of these types of manipulations. e.g.

x^4 = 25
x^4 - 25 = 0
(x^2 - 5)(x^2 + 5) = 0
 
  • #11
To continue from where Hurkyl left off:

x^{4}-25=(x-\sqrt{5})(x+\sqrt{5})(x-i\sqrt{5})(x+i\sqrt{5})=0

I think the solutions are obvious.

Daniel.
 
  • #12
dextercioby said:
To continue from where Hurkyl left off:

x^{4}-25=(x-\sqrt{5})(x+\sqrt{5})(x-i\sqrt{5})(x+i\sqrt{5})=0

I think the solutions are obvious.
Obvious? Probably not to jai6638
 
  • #13
And how did u know that??Did u talk to him?? :wink:

Daniel.
 
  • #14
i got it... thanks much! :)
 
  • #15
dextercioby said:
And how did u know that??Did u talk to him?? :wink:

Daniel.
Well, he(she?) didn't know how to find the zeros of x^4-25. From that I can infer that you statement might not be very obvious to him(her?).
 
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