Question

help please !!!
simplify completley:

Answer 1

\[\frac{ t ^{4}-1 }{ t ^{3}+t ^{2}+t+1 }\]
bottom equation is t^3 + t^2 + t + 1

Answer 2

Do you see anything that can be factored in the numerator?

Answer 3

can it be \[{( t ^{2}-1)(t ^{2}+1) }\]

Answer 4

Yes it can

Answer 5

Do you see another application of the same thing within the numerator?

Answer 6

no, i do not. can you please help me

Answer 7

Alright, so the first time you used difference of squares to factor \[t^4-1 = (t^2-1)(t^2+1)\]If you use difference of squares again, on the first factor(t^2-1), you will get a factored version of the numerator.\[t^4-1 = (t^2-1)(t^2+1) = (t-1)(t+1)(t^2+1)\]From here, you just need to factor the denominator so that you can cancel factors and simplify.

Answer 8

Do you see any way of factoring the denominator?

Answer 9

no, i do not know how to factor four "numbers".

Answer 10

lol, have you learned about the factor theorem?

Answer 11

or synthetic divison for that matter?

Answer 12

not that i recall, no

Answer 13

hmm, well the factor that I noticed was when I factored t^2 from the first two terms.\[t^3 + t^2 + t + 1 = t^2(t+1) + t+ 1\]What might be more noticeable now is that factoring by grouping is an option now.\[ t^2(t+1) + (t+1) = (t^2+1)(t+1)\]

Answer 14

so the answer would be t-1
because the t^2+1 cancel out from both top and bottom and the t+1 cancel out as well, so whats left is t-1.

Answer 15

exactly

Answer 16

okay, thank you so much for your help and time. i really appreciate it!(:

Answer 17

np