Advanced polynomial problem

Algebra Level 3

Let $P_n (x)$ be a polynomial such that $P_n (x) = P_{n-1} (x-n)$ .

Given that $P_0 (x)= x^{90} - x^{89} + x^{88} - x^{87} + \cdots + 1$ , and $P_{10} (x) = P_0 (x-k)$ , find $k$ .

54 44 55 45

1 solution

Anuj Soni
Apr 10, 2016

According to the question, P0(x) = x^90 - x^89 + x^88 -....... - x +1 now, P1(x) = P0(x-1) so, P1(x) = (x-1)^90 -(x-1)^89 ..... - (x-1) +1 and repeating these steps to P10(x) we get P10(x) = (x-55)^90 - (x-55)^89..... -(x-55) +1 therefore P10(x) = P0(x-55) hence k = 55

Advanced polynomial problem

1 solution

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