Find the zeros of the function. State the multiplicity of any multiple zeros. y=x^4-8x^2+16

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Find the zeros of the function. State the multiplicity of any multiple zeros. y=x^4-8x^2+16

Mathematics
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have u tried this ?
ya im supposed to find the GCF but i don't think there is one so i dont know what to do
GCF ? first put x^2=y what u get ?

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why would x^2 equal y
oh, it was just for simplification, u can put x^2 =t .... or z ..
i didn't mean that y on left side...
square root both sides
to find zeros of y=x^4-8x^2+16 u put y=0 x^4-8x^2+16=0 now if we put x^2 = t t^2 -8t +16 = 0 this is quadratic in t, can u solve it ?
so just use the quadratic formula
or u can factor it .
the answer is -4 or 4 but the answer is supposed to be -2 or 2
how do i factor it
yes, because x^2 is 4,4
t^2-8t+16 = 0 t^2 -4t -4t +16 = 0
ok?
what u got t as ?
im sorry what?
t^2 -4t -4t +16 = 0 can u solve for t ?
no thats why i asked
t^2 -4t -4t +16 = 0 factor out t from first 2 terms , what u get ?
i mean t(t-8)+16=0
t^2 -4t -4t +16 = 0 from this.... factor out t from 1st 2 terms....
thats what i thought i just did
u factored t from t^2 -8t +16 = 0 i asked u to factor out t from t^2 -4t -4t +16 = 0
t(t-4-4)+16=0
i mentioned 'from first two terms' ...
t(t-4-4t)+16=0
t^2-4t-4t+16=0 t(t-4) -4 (t-4) = 0 did u get this ?
oh ok that makes sense
t(t-4) -4 (t-4) = 0 (t-4)(t-4) =0 i factored t-4 from that, got this ?
how did u get rid of the t and -4 ouside the parentheses
i did not get rid of that...its in the 2nd brackets '()' ... (t-4)
if u are getting confused let a=t-4 t(t-4) -4 (t-4) = 0 ta-4a = 0 a(t-4)=0 now put back a= t-4 (t-4)(t-4) = 0 got this ?
ok
so, t=4,4 which gives u x^2 = 4, 4
so whats x = ?
x=2 and -2
x=+or - the square root of 4
yes thats correct...x=2,-2 for x^2 =4 now since x^2 = 4,4 ....u get x= 2,2,-2,-2 that is a root =2 with muliplicity 2 and another root =-2 with multiplicity = 2

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