Answer to Question #270262 in Algebra for pica

Question

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).

Answer 1

Let $q_1(x), q_2(x), q_3(x)$ 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

$f(x) = (x+1)q_1(x) + 5 \;\;\;(1) \\ f(x) = (x-1)q_2(x) +1 \;\;\;(2) \\ f(x) = (x-2)q_3(x) +20 \;\;\;(3)$

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

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

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

$f(x) = (x+1)(x-1)(x-2)q_n(x) + (ax^2 +bx+c) \\ f(-1) = a-b+c \\ a-b+c = 5 \;\;\;(4) \\ f(1) = a+b+c \\ a+b+c=1 \;\;\;(5) \\ f(2) = 4a+2b+c \\ 4a+2b+c = 20 \;\;\;(6)$

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

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

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

$4a+2b+c = 20 \\ 2a+2b+2c = 2 \\ 2a-c = 18$

Adding (7) and (8)

$a+c = 3 \\ 2a-c = 18 \\ 3a=21 \\ a = 7$

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

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

Remainder is $7x^2 -2x -4$