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)}


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Priyansh Bajaj