What is the completely factored form of t4 − 16?

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What is the completely factored form of t4 − 16?

Mathematics
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ok did u try it?
Yes. I got (t2-4)(t2+4)
Thats correct :)

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Other answers:

ok thanks:)
Wait! No it's not. I got it wrong:(
(t2+4)(t+2)(t−2)
i was thinking both of them are the same thing
Ohhh okay :)
what was the right answer?
Yes:)
:) welcome to openstudy :)
haha thanks :)
Factoring means factoring completely. When you have to factor a polynomial of two terms, look for the difference of squares or the sum or difference of cubes. The first step in factoring, though, is always to look for a common factor. In this case there is no common factor. Then you have two terms. t^4 is obviously not a cube, so look at the difference of two squares. Sure enough, \(t^4 - 16\) can be written as \( (t^2) - 4^2 \) which clearly looks like the difference of two squares.
The difference of squares factors like this: \(a^2 - b^2 = (a + b)(a - b) \) Your expression factors similarly: \((t^2)^2 - 4^2 = (t^2 + 4)(t^2 - 4) \) Now you need to look at each factor to see if it can be factored further. \(t^2 + 4\) is a sum of squares. It can't be factored. Now we look at \(t^2 - 4\) which is a difference of squares and can be factored. \(t^4 - 16 = (t^2)^2 - 4^2 = (t^2 + 4)(t^2 - 4) = (t^2 + 4)(t + 2)(t - 2) \) Finally we look at the new factors, t + 2 and t - 2, and we see that they can;t be factored any further, so our answer above is fully factored.

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