Question

If \(x^{100} + 1 \) is divided by \(x^2 - 1\), then what is the remainder?

Answer 1

Polynomial remainder theorem doesn't work here, unfortunately. Does it?

Answer 2

\( \color{Black}{\Rightarrow x^2 - 1 = (x + 1)(x - 1) }\)

Answer 3

this may help u http://in.answers.yahoo.com/question/index?qid=20120328050459AASixcw

Answer 4

I need to work on my own, and I need step-by-step guidance.

Answer 5

x^2-1=(x+1)(x-1) .. first step:)

Answer 6

i got something please wait

Answer 7

x^2-1=0
x^2=1
x=1

x^(100)+1=1^(100)+1=2

Answer 8

I see.

Answer 9

remainder theorem says that .. if p(x) is divided by p(a) then

Answer 10

sorry leave that :

let x^2-1=0

Answer 11

then x = ?

Answer 12

How can we even divide a number by 0?

Answer 13

no dont divide .. equate that as 0

\[\large{x^2-1=0}\] x =?

Answer 14

\(x = \pm 1\)

Answer 15

Now we can find \(p(1)\)

Answer 16

right now put that in p(x) = \(\large{x^{100}+1}\) ..p(1)=? remainder

Answer 17

I do know the remainder theorem for \(p(x) \div x - k\), but I didn't know how to apply that here.

Why do we equate with 0?

Answer 18

we generally : suppose x-k=0 that is x =k and then remainder = p(k) ..

Answer 19

Oh, I see.

Answer 20

Thank you!

Answer 21

did that seriously helped you ..

i dont think i did

Answer 22

I think you needed the proof for that thingy ?

Answer 23

No, I understand it fully now!
I do have it.

Answer 24

ok just for reference : ncert IX chapter 2 has explanation for this theorem

Answer 25

well nice to hear that u got it .. :)

Answer 26

I do have it :P

Answer 27

:P good

Answer 28

Thanks again!

Answer 29

no problem.. have to go now bbye

Answer 30

Bye! :)