Question

The remainder when the polynomial 1 + x2 + x4 + x6 +...+ x22 is divided by
1+x+ .x2 + x3 + ... +811 is
A. O
B. 2.
C. It x2 + x4 + ... + x 10
D. 2(1+12 + x4 + ... + x10 )​

Answer 1

Remainder = 2    if 1 + x² + x⁴ + x⁶ +...+ x²² is divided by 1 + x + x² + x³ +...+ x¹¹

Step-by-step explanation:

1 + x² + x⁴ + x⁶ +...+ x²²

this is an GP

a = 1

r = x²

n = 12

Sum = 1 ( x²⁴ - 1)/(x² - 1)

1 + x + x² + x³ +...+ x¹¹

this is an GP

a = 1

r = x

n = 12

Sum = 1 ( x¹² - 1)/(x - 1)

( x²⁴ - 1)/(x² - 1) /  ( x¹² - 1)/(x - 1)

=( x²⁴ - 1)  / ( x¹² - 1)(x + 1)

= ( x¹² + 1)( x¹² - 1)/( x¹² - 1)(x + 1)

=  ( x¹² + 1)/(x + 1)

Putting x = -1

we get remainder = 2

Remainder = 2

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