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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

The sum of the series $\sum\limits_{r = 0}^n {(-1)^r \, ^nC_r \left( \frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} + \dots + m \text{ terms} \right)}$ is

The coefficient of $x^{2012}$ in the expansion of $(1-x)^{2008}(1+x+x^2)^{2007}$ is equal to

If $T_r = ^{2016}C_r x^{2016-r}$ for $r = 0, 1, 2, \dots, 2016$,then $(T_0 - T_2 + T_4 - \dots + T_{2016})^2 + (T_1 - T_3 + T_5 - \dots - T_{2015})^2$ is equal to-

If ${T_0}, {T_1}, {T_2}, \dots, {T_n}$ represent the terms in the expansion of ${(x + a)^n}$,then $({T_0} - {T_2} + {T_4} - \dots)^2 + ({T_1} - {T_3} + {T_5} - \dots)^2 = $

The sum of the series $2 \times 1 \times {}^{20}C_4 - 3 \times 2 \times {}^{20}C_5 + 4 \times 3 \times {}^{20}C_6 - 5 \times 4 \times {}^{20}C_7 + \dots + 18 \times 17 \times {}^{20}C_{20}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module