The remainder when the polynomial 1+x^2+x^4+x | vedclass

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(D) Let $P(x) = 1+x^2+x^4+\ldots+x^{22} = \frac{(x^2)^{12}-1}{x^2-1} = \frac{x^{24}-1}{x^2-1}$.Let $Q(x) = 1+x+x^2+\ldots+x^{11} = \frac{x^{12}-1}{x-1

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$2(1+x^2+x^4+\ldots+x^{10})$

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(D) Let $P(x) = 1+x^2+x^4+\ldots+x^{22} = \frac{(x^2)^{12}-1}{x^2-1} = \frac{x^{24}-1}{x^2-1}$.Let $Q(x) = 1+x+x^2+\ldots+x^{11} = \frac{x^{12}-1}{x-1}$.We can write $P(x) = \frac{(x^{12}-1)(x^{12}+1)}{(x-1)(x+1)} = Q(x) \cdot \frac{x^{12}+1}{x+1}$.Using polynomial division,$\frac{x^{12}+1}{x+1} = \frac{x^{12}-1+2}{x+1} = \frac{(x+1)(x^{11}-x^{10}+x^9-\ldots-1)+2}{x+1} = (x^{11}-x^{10}+x^9-\ldots-1) + \frac{2}{x+1}$.Thus,$P(x) = Q(x) \cdot (x^{11}-x^{10}+x^9-\ldots-1) + \frac{2 Q(x)}{x+1}$.Since $Q(x) = \frac{x^{12}-1}{x-1} = (x+1)(x^{10}-x^8+x^6-x^4+x^2-1)$ is not quite right,let's use $Q(x) = (1+x)(1+x^2+x^4+x^6+x^8+x^{10})$.Then $\frac{2 Q(x)}{x+1} = 2(1+x^2+x^4+x^6+x^8+x^{10})$.Therefore,the remainder is $2(1+x^2+x^4+x^6+x^8+x^{10})$.

The remainder when the polynomial $1+x^2+x^4+x^6+\ldots+x^{22}$ is divided by $1+x+x^2+x^3+\ldots+x^{11}$ is

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