SOLUTION: Good evening One more question. I think I have the answer yet i'm only 60% sure. Find all the zeros of f(x) if f(x) =(x^2-4)^3. Identify any multiple zeros. f(x)=0 (x^2-4

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Good evening One more question. I think I have the answer yet i'm only 60% sure. Find all the zeros of f(x) if f(x) =(x^2-4)^3. Identify any multiple zeros. f(x)=0 (x^2-4      Log On


   

Question 257025: Good evening
One more question. I think I have the answer yet i'm only 60% sure.
Find all the zeros of f(x) if f(x) =(x^2-4)^3. Identify any multiple zeros.
f(x)=0
(x^2-4)^3=0
F(4)=(4^2-4)^3=48-12=36-36=0
F(0)=(0^2-4)^3=0-12=-12+12=0
Answer: {4, (triple zero)}
Thank you much
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
The original question was
f(x) =(x^2-4)^3.
step 1 - set = 0 to get
(i) f%28x%29+=%28x%5E2-4%29%5E3+=+0
take a cube root to get
(ii) f%28x%29+=%28x%5E2-4%29+=+0
Factoring, we get
(iii) %28x-2%29%28x%2B2%29+=+0
x = 2 and x = -2
It turns out that
x = 2 with multiplicity 3
x = -2 with multiplicity 3