Question #270262

When a polynomial is divided by x + 1, the remainder is 5. When the same polynomial is divided by x – 1, the remainder is 1. When the same polynomial is divided by x – 2, the remainder is 20. Determine the remainder when the polynomial is divided (x + 1)(x – 1)(x – 2).


1
Expert's answer
2021-11-29T02:29:23-0500

Let q1(x),q2(x),q3(x)q_1(x), q_2(x), q_3(x) be the quotient when polynomial f(x) is divided by (x+1), (x-1) and (x-2) respectively.

By Division Algorithm

Dividend = Divisor ×\times Quotient + Remainder

f(x)=(x+1)q1(x)+5      (1)f(x)=(x1)q2(x)+1      (2)f(x)=(x2)q3(x)+20      (3)f(x) = (x+1)q_1(x) + 5 \;\;\;(1) \\ f(x) = (x-1)q_2(x) +1 \;\;\;(2) \\ f(x) = (x-2)q_3(x) +20 \;\;\;(3)

Put x = -1 in equation (1), x=1 in equation (2) and x = 2 in equation (3), we get

f(1)=5f(1)=1f(2)=20f(-1) = 5 \\ f(1) = 1 \\ f(2) = 20

Let ax2+bx+cax^2 +bx +c be the remainder when polynomial f(x) is divided by (x+1)(x-1)(x-2) and quotient = q_n(x)

f(x)=(x+1)(x1)(x2)qn(x)+(ax2+bx+c)f(1)=ab+cab+c=5      (4)f(1)=a+b+ca+b+c=1      (5)f(2)=4a+2b+c4a+2b+c=20      (6)f(x) = (x+1)(x-1)(x-2)q_n(x) + (ax^2 +bx+c) \\ f(-1) = a-b+c \\ a-b+c = 5 \;\;\;(4) \\ f(1) = a+b+c \\ a+b+c=1 \;\;\;(5) \\ f(2) = 4a+2b+c \\ 4a+2b+c = 20 \;\;\;(6)

Adding equation (4) and (5), we get

2a+2c=6a+c=3      (7)2a+2c= 6 \\ a+c = 3 \;\;\;(7)

Multiplying equation (5) by 2 and subtracting it from equation (6)

4a+2b+c=202a+2b+2c=22ac=184a+2b+c = 20 \\ 2a+2b+2c = 2 \\ 2a-c = 18

Adding (7) and (8)

a+c=32ac=183a=21a=7a+c = 3 \\ 2a-c = 18 \\ 3a=21 \\ a = 7

Using a in equation (7), c = -4

Using a and c in equation (5), b = -2

Remainder is 7x22x47x^2 -2x -4


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