Respuesta :
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:
I have a lot of factors mentioned at the end of this explanation.
Step-by-step explanation:
[tex]t^4-81[/tex] is a difference of squares since we can write it as [tex](t^2)^2-(9)^2[/tex].
A difference of squares, [tex]a^2-b^2[/tex], can be factored as [tex](a-b)(a+b)[/tex].
[tex]t^4-81[/tex]
[tex](t^2)^2-(9)^2[/tex]
[tex](t^2-9)(t^2+9)[/tex]
We see another difference of squares in this factored form.
I'm speaking of [tex]t^2-9[/tex].
This can be rewritten as [tex](t)^2-(3)^2[/tex].
Let's factor it now.
[tex]t^2-9[/tex]
[tex](t)^2-(3)^2[/tex]
[tex](t-3)(t+3)[/tex]
So [tex]t^4-81=(t^2-9)(t^2+9)=(t-3)(t+3)(t^2+9)[/tex].
Now [tex]t^2+9[/tex] can also be factored if you invite all complex numbers into play.
You will need [tex]i^2=-1[/tex] here.
[tex]t^2+9[/tex]
[tex]t^2-(-9)[/tex]
[tex]t^2-(9i^2)[/tex]
[tex]t^2-(3i)^2[/tex]
Now it is a difference of squares and we can do as we have been doing with the other factors:
[tex](t-3i)(t+3i)[/tex]
So the complete factored form of [tex]t^4-81[/tex] is:
[tex](t-3)(t+3)(t-3i)(t+3i)[/tex].
So here are some things that you could say is a factor of [tex]t^4-81[/tex]:
[tex]1[/tex]
[tex]-1[/tex]
[tex]t-3[/tex]
[tex]-(t-3)[/tex]
[tex]-t+3[/tex] (same as one before; just a rewrite)
[tex]t+3[/tex]
[tex]-(t+3)[/tex]
[tex]-t-3[/tex] (same as one before; just a rewrite)
[tex]t^2-9[/tex]
[tex]-(t^2-9)[/tex]
[tex]-t^2+9[/tex] (same as one before; just a rewrite)
[tex]t^2+9[/tex]
[tex]-(t^2+9)[/tex]
[tex]-t^2-9[/tex] (same as one before; just a rewrite)
[tex]t-3i[/tex]
[tex]-(t-3i)[/tex]
[tex]-t+3i[/tex] (same as one before; just a rewrite)
[tex]t+3i[/tex]
[tex]-(t+3i)[/tex]
[tex]-t-3i[/tex] (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.
