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Find the remainder when (x+1)^n is divided by (x-1)^3. Askiitians tutors please help. No one replied when i posted this question before. This question is of rmo. Please give the solution.

Find the remainder when (x+1)^n is divided by (x-1)^3.
Askiitians tutors please help. No one replied when i posted this question before. This question is of rmo.
Please give the solution.

Grade:11

6 Answers

Pratik Tibrewal
askIITians Faculty 37 Points
7 years ago
If any polynomial P(x) is divided by Q(x) and if it generates remainder f(x), and quotient is g(x),
then P(x) = Q(x) . g(x) + f(x)
therefore (x+1)^n = (x-1)^3 . g(x) + f(x)
here f(x) would be a quadratic equation, lets say ax^2 + bx + c;
so: (x+1)^n = (x-1)^3 . g(x) + ax^2 + bx + c;
Put x = 1; we get: 2^n = a + b + c
now differentiate, the equation once and put x =1;
n . 2^(n-1) = 2a + b
differentiate it again; and put x = 1;
n.(n-1).2^(n-2) = 2a

Hence the remainder is:R(x) = [2^(n - 3)] * [n(n - 1)(x^2) + (4n - (2n^2 - 2n))x + (n(n - 1) - 4n + 8]

Thanks and Regards,
Pratik Tibrewal,
askiitians faculty,
BTech
Shrey
49 Points
7 years ago
Thank you sir!
Pushkar Aditya
71 Points
7 years ago
Sorry for inconvenience Sir. The answer is n(n-1)*2^n-3*x^2-2^n-2*x(n^2-3n)+2^n-3(n^2-5n+8) Please see to it
Shrey
49 Points
7 years ago
Sorry for inconvenience. the answer is: n(n-1)*2^n-3*x^2-2^n-2*x(n^2-3n)+2^n-3(n^2-5n+8) This is the correct answer sir,please check once again.
GAURAV SINGH
askIITians Faculty 36 Points
7 years ago
When we divide polynomial P(x) by quotient Q(x) and get remainder R(x), this means:
P(x) = D(x)*Q(x) + R(x)
Order of the polynomials: P is of order n, Q of order 3, D of order (n - 3) and R of order 2.
Let R(x) = ax^2 + bx + c:
(x + 1)^n = D(x)*(x - 1)^3 + (ax^2 + bx + c)
We can eliminate the term with unknown polynomial D(x) by substituting x = 1:
2^n = a + b + c
Differentiate both sides and substituting x = 1 again to obtain another equation:
n*[(x + 1)^(n - 1)] = g(x - 1) + (2ax + b) where g is some function of (x - 1) so when x = 1, g = 0.
n*(2^(n - 1)) = 2a + b
Differentiate original equation twice:
n*(n - 1)*[(x + 1)^(n - 2)] = h(x - 1) + 2a; h is another function where h = 0 when x = 1.
When x = 1, n*(n - 1)*(2^(n - 2)) = 2a.
Hence we have 3 equations for 3 unknowns a, b and c:
U = 2^n = a + b + c
V = n*(2^(n - 1)) = 2a + b
W = n*(n - 1)*(2^(n - 2)) = 2a
giving
a = W/2
b = V - W
c = U - (a + b) = W/2 - V + U
Then R(x) = ax^2 + bx + c
R(x) = (W/2)(x^2) + (V - W)*x + (W/2 - V + U)
To simplify this, we subsitute:
W = [2^(n - 3)]*[2n(n - 1)]
W/2 = [2^(n - 3)]*[n(n - 1)]
V = n*(2^(n - 1)) = [2^(n - 3)]*[4n]
U = 2^n = [2^(n - 3)]*[8]
R(x) = [2^(n - 3)] * [n(n - 1)(x^2) + (4n - (2n^2 - 2n))x + (n(n - 1) - 4n + 8]
R(x) = [2^(n - 3)] * [n(n - 1)(x^2) + (- 2n^2 + 6n)x + (n^2 - 5n + 8)]

Thanks & Regards
Gaurav Singh
askIITians Faculty
Shrey
49 Points
7 years ago
This is the perfect answer.Thanks!!!!!

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