Factor the following as if it were a trinomial

9t^1/1-25

I am sure it was not worded like that in your textbook.

hmmm...that's how it's worded.

Does this help?

(9t^1/1)-25

I know most of the answer is (3t+5)(3t-5). What I don't understand is the exponent of 1/1. How would that be factored in the equation?

nope,

the 1/1 part is simply 1 , so 9t^1/1 simply becomes 9t
so you have just 9t - 25 , which does not factor

The computer says that 9t-25 isn't correct. Any thoughts?

Must be the way you typed the original question.

If the answer is (3t+5)(3t-5) then the original question must have been

9t^2 - 25 , which is not even close to what you typed.

No, that's not it either. The directions say to factor as if it were a trinomial. I'm probably typing the actual problem wrong. I'll try to use words in the problem and see if that helps.

9t with the exponent of 1/1 minus 25

And as I said before, 1/1 is merely 1, so why even have the exponent ?

If it were the way I suspect as 9t^2 - 25
you could write it as a trinomial by inserting 0t and now you have a trinomial

9t^2 + 0t - 25
= (3t+5)(3t-5)

but you claim that is not right.

I don't think I can be of any more help if you insist that the question was as you typed it.

Thanks for trying to help. The answer was (3t^1/2+5)(3t^1/2 -5).

Well yeahh!

That would make it
(3√t + 5)(3√t - 5) = 9t - 25 , and the
9t - 25 is what I had pointed out to you above.

Usually, the instruction to "factor" implies that we restrict ourselves to rational numbers.
If you allow irrationals, like the above, or even imaginary numbers, then anything goes.

9t^(1/1) - 25 is just a very silly way to write 9t-25