What is the remainder when x^23 is divided by x^2-3x+2 ?

Dear Vignesh,

Solution:- x^2 - 3x + 2 can be written as (x-1)(x-2)

Also, x^23 can be written as [2(x-1) - (x-2)]²³, which on Binomial expansion will give-

[2(x-1) - (x-2)]²³ = [2(x-1)]²³ + (-1) ²³C₁[2(x-1)]²² [(x-2)] + (-1)^{2 23}C₂[2(x-1)]²¹ [(x-2)]² +.....(-1)^{22 23}C₂₂[2(x-1)] [(x-2)]²² + (-1)^{23 23}C₂₃ [(x-2)]²³

As we can see, all the terms except first and last contains atleast one (x-1) and (x-2).

So, the terms that are not divisible by (x-1)(x-2) are [2(x-1)]²³ and (-1)^{23 23}C₂₃ [(x-2)]²³

Remainder = { [2(x-1)]²³ + (-1)^{23 23}C₂₃ [(x-2)]²³ } /(x-1)(x-2)

              = { [2(x-1)]²² /(x-2)} - {[(x-2)]²² /(x-1)}


Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.


All the best Vignesh!!!



Regards,

Askiitians Experts

Priyansh Bajaj

Approved