NTSE Stage 2 SAT 2019 Exam Question Paper

Question : 57 of 100

Marks: +1, -0

(x+1)^4

is divided by

(x-1)^3

Then the value of the remainder at

x=1

Solution:

(x+1)^{4}=x^{4}|+4x^{3}+6x^{2}+4x+1

(x-1)^{3}=x^{3}-3x^{2}+3x-1

x^{3}-3x^{2}+3x-1\;\;)\;\;x^{4}+4x^{3}+6x^{2}+4x+1\;\;(\;\;x+7

x^4-3x^{3}+3x^{2}-x

-\;+\;\;\;\;-\;\;\;\;\;+

––––––––––––––

7x^{3}+3x^{2}+5x+1

7x^{3}-21x^{2}+21x-7

-\;\;\;+\;\;\;\;\;\;\-\;\;\;\;\;\;+

––––––––––––––––

24x^{2}-16x+8

x=1

, remainder

=24-16+8=16

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