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

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

Grade:8

1 Answers

Priyansh Bajaj AskiitiansExpert-IITD
30 Points
10 years ago

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 23C2[2(x-1)]21 [(x-2)]2 +.....(-1)22 23C22[2(x-1)] [(x-2)]22 + (-1)23 23C23 [(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 23C23 [(x-2)]23

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

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


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All the best Vignesh!!!



Regards,

Askiitians Experts

Priyansh Bajaj

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