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astraxan [27]
3 years ago
14

Find the remainder for (x^10 + x^9 +...+x+1) divided by (x^2 - 1)

Mathematics
1 answer:
Assoli18 [71]3 years ago
3 0

Since x^2-1=(x-1)(x+1), by the remainder theorem we have

\dfrac{p(x)}{x-1} = q(x) + \dfrac{p(1)}{x-1} = q(x) + \dfrac{11}{x-1}

where p(x) = x^{10}+x^9+\cdots+x+1.

Then

\dfrac{p(x)}{x^2-1} = q^*(x) + \dfrac{q(-1)}{x+1} + \dfrac{11}{x^2-1} = q^*(x) + \dfrac{q(-1)(x-1) + 11}{x^2-1}

The only missing piece is <em>q(x)</em>, which we can get through usual polynomial division:

\dfrac{x^{10}+x^9+\cdots+x+1}{x-1} = \underbrace{x^9 + 2x^8 + 3x^7 + \cdots + 9x + 10}_{q(x)} + \dfrac{11}{x-1}

so that

q(-1) = (-1)^9 + 2(-1)^8 + \cdots + 9(-1) + 10 = 5

Then the remainder we want is

\dfrac{p(x)}{x^2-1} = q^*(x) + \dfrac5{x+1} + \dfrac{11}{x^2-1} = q^*(x) + \dfrac{5(x-1)+11}{x^2-1} = q^*(x) + \dfrac{\boxed{5x+6}}{x^2-1}

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