Question

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

Answer 1

Answer:

±3

Step-by-step explanation:

t^4 - 81 = 0

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

t^2 = 9 = ±3

t^2 = -9 no solution

Answer 2

Answer:

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

Step-by-step 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.