The discussion centers on simplifying the polynomial expression \(125t^3 + 75t^2 + 50t + 4\) to \(65t + 5\) under modulo 25. Participants clarify that since \(125 \equiv 0 \ (\text{mod} \ 25)\), the terms involving \(125t^3\), \(75t^2\), and \(50t\) vanish, leaving \(65t + 5\) as the simplified result. The key takeaway is understanding how modular arithmetic reduces polynomial expressions by eliminating terms that are multiples of the modulus.
PREREQUISITESStudents studying algebra, particularly those focusing on modular arithmetic and polynomial simplifications, as well as educators looking for examples to illustrate these concepts.
125\equiv 0\ (\!\!\!\!\mod25)Instinctlol said:Homework Statement
I am looking at an example from my book and I have no clue how this is done.
Homework Equations
I solved the left hand side and got this
125t3+75t2+50t+4
How did they reduce it to 65t+5?
The Attempt at a Solution
http://i48.tinypic.com/2v2gpap.jpg
SammyS said:125\equiv 0\ (\!\!\!\!\mod25)
etc.
Instinctlol said:I don't understand please clarify
micromass said:Do you understand what (mod 25) means?
Instinctlol said:I don't understand where he got the 125 and what happened to the t
If \ 125\equiv 0\ (\!\!\!\!\mod25)\, then \ 125t^3\equiv 0t^3\ (\!\!\!\!\mod25)Instinctlol said:I don't understand where he got the 125 and what happened to the t
SammyS said:If \ 125\equiv 0\ (\!\!\!\!\mod25)\, then \ 125t^3\equiv 0t^3\ (\!\!\!\!\mod25)
You do understand that \ 125\equiv 0\ (\!\!\!\!\mod25)\,,\ don't you?
Anyway ...Instinctlol said:That means 25|125, sorry I guess the variables just confuses me.
SammyS said:Anyway ...
Do you now understand why \displaystyle \ \ 250 t^3+150 t^2+65 t+ 5 \equiv 65t+5 ~\text{(mod 25)}\ ?
Yes, I get \displaystyle \ \ 2(1+5t)^3+7(1+5t)-4= 250 t^3+150 t^2+65 t+ 5 \ .
Well, the remainder of 65t+5 is the same as the remainder of 250t3+150t2+65t+5 when divided by 25 .Instinctlol said:Is it because 25|250 and 25|150 so the remainder is 65t+5?