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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
10 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
10 years ago
Thank you sir!
Pushkar Aditya
71 Points
10 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
10 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
10 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
10 years ago
This is the perfect answer.Thanks!!!!!

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