what r the factors of t^4-81?

Answer 1

Some of the above are partially correct.

If you restrict yourself to the reals then there are only 2.

you can factor as

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

But as you may or may not know if you allow complex roots then a nth degree polynomial has exactly n roots.

The latter part can be factored with the imaginary roots giving finally

(t-3)(t+3)(t-3i)(t+3i)

Answer 2

3 ^ 4 = 81 so t = 3

Answer 3

this is a really interesting question. the one algebric property that this question manipulates is (a + b)(a - b) = a^2 - b^2.
to use the above mention property, we have to try to tweak the question a little so that we can get something close to a^2 - b^2.

t^4 = (t^2)^2
81 = 9^2

therefore by implementing the algebric property above we can deduce that:
t^4-81 = (t^2 - 9)(t^2 + 9)

but that's not all. if you noticed, (t^2 - 9) also follows the algebric property.
hence: (t^2 - 9) = (t + 3)(t - 3)

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

*number guy is incorrect by the way*

if you are wondering how i knew that i needed to use the (a + b)(a - b) = a^2 - b^2 property, it's because of practice, practice and practice. after a while you just seem to know which ones to use and which ones not to. you'll get the hang of it soon as long as you drill yourself. good luck and have fun with math.

Answer 4

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

to take this further you would get

(t - 3)(t + 3)(t + 3i)(t - 3i)

Answer 5

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

Answer 6

(t^2+9)(t+3)(t-3)

Answer 7

t^-77