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Q: What is the factorization of T power 4 - 81?

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t^4 - 81 = (t^2)^2 - (3^2)^2 = (t^2 - 3^2)(t^2 + 3^2) = (t - 3)(t + 3)(t^2 + 9)

16t(t + 4) is the factorization Usually it's set = to 0 16t (t+4) =0 So, either 16t = 0 or t + 4 = 0 t = 0 or t = -4

T to the fourth power is already fully simplified unless you have a value for "t."

t with a small 4 to the top right of it

1

You can take a 5s2t out of all of that, leaving yourself with 3s4t3 + 19s3t2 + 7s2t4 - 7

The prime factorization for 68 is: 2 x 2 x 17

2(t + 10)

(t - 3)(t + 3)(t2 + 9)

t4-81 is a difference of 2 squares and can be written as (t2-9)(t2+9) t2+9 can't be further factorised but t2-9 is a difference of 2 squares again and can be factorised to (t+3)(t-3) so the factors of t4-81 are :(t2+9)(t+3)(t-3) Hope this helps :-) I believe the answer you are looking for is (t - 3)(t + 3)(t 2 + 9)

16 + 49 = 65

Edwin T. Layton died on April 12, 1984 at the age of 81.

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