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 q(x), 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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