Question

Which of the following are the factors of t^4 -81

Answer 1

±3

Explanation:

t^4 - 81 = 0

(t^2 -9) (t^2 + 9) = 0

t^2 = 9 = ±3

t^2 = -9 no solution

Answer 2

I have a lot of factors mentioned at the end of this explanation.

Explanation:

`t^4-81` is a difference of squares since we can write it as
`(t^2)^2-(9)^2`.

A difference of squares,
`a^2-b^2`, can be factored as
`(a-b)(a+b)`.

`t^4-81`

`(t^2)^2-(9)^2`

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

We see another difference of squares in this factored form.

I'm speaking of
`t^2-9`.

This can be rewritten as
`(t)^2-(3)^2`.

Let's factor it now.

`t^2-9`

`(t)^2-(3)^2`

`(t-3)(t+3)`

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

Now
`t^2+9` can also be factored if you invite all complex numbers into play.

You will need
`i^2=-1` here.

`t^2+9`

`t^2-(-9)`

`t^2-(9i^2)`

`t^2-(3i)^2`

Now it is a difference of squares and we can do as we have been doing with the other factors:

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

So the complete factored form of
`t^4-81` is:

`(t-3)(t+3)(t-3i)(t+3i)`.

So here are some things that you could say is a factor of
`t^4-81`:

`1`

`-1`

`t-3`

`-(t-3)`

`-t+3` (same as one before; just a rewrite)

`t+3`

`-(t+3)`

`-t-3` (same as one before; just a rewrite)

`t^2-9`

`-(t^2-9)`

`-t^2+9` (same as one before; just a rewrite)

`t^2+9`

`-(t^2+9)`

`-t^2-9` (same as one before; just a rewrite)

`t-3i`

`-(t-3i)`

`-t+3i` (same as one before; just a rewrite)

`t+3i`

`-(t+3i)`

`-t-3i` (same as one before; just a rewrite)

There are other ways to write some of these hopefully you can catch them on your own in your choice if they so occur.