Question

the remainder R obtained by dividing x 100 by x2-3x+2 is the polynomial of degree less than 2 .
then R may be written as:
(A)2100-1
(B)2100 (x-1)-(x-2)
(C)2100 (x-3)
(D) x(2100-1)+2(299-1)

Answer 1

Answer:

(B)2100 (x-1)-(x-2)

Step-by-step explanation:

Let  p(x) = x100 and q(x) = x2 – 3x + 2

= x2 – 2x – x + 2 = x(x– 2) – 1(x – 2)

 = (x – 2)(x – 1)

Let Q(x) and r(x) be the quotient and remainder when p(x) is divided by q(x),

Hence by division algorithm, we have

p(x) = q(x) Q(x) + r(x)

That is,  x100 = (x – 2)(x – 1)Q(x) + r(x)

Let r(x) = (ax + b) where 0 < deg r(x) < 3

x100 = (x – 2)(x – 1)Q(x) + (ax + b) → (1)

Put x = 1 in equation (1), we get

1 = 0 + (a + b)

Hence (a + b) = 1 → (2)

Now put x = 2 in equation (1), we get

2100 = 0 + (2a + b)

2a + b = 2100 → (3)

Solving (2) and (3), we get

a = (2100 – 1) and b = 2 – 2100

Therefore the remainder = (ax + b) = (2100 – 1)x + (2 – 2100)

= (B) 2100 (x-1)-(x-2)