From the formula for a geometric series,
\[p(x) = 1 + x^2 + x^4 + x^6 + \dots + x^{22} = \frac{x^{24} - 1}{x^2 - 1}.\]Likewise,
\[q(x) = 1 + x + x^2 + x^3 + \dots + x^{11} = \frac{x^{12} - 1}{x - 1}.\]At first, it may look like we can write $p(x)$ as a multiple of $q(x)$:
\[\frac{x^{24} - 1}{x^2 - 1} = \frac{x^{12} - 1}{x - 1} \cdot \frac{x^{12} + 1}{x + 1}.\]Unfortunately, $\frac{x^{12} + 1}{x + 1}$ is not a polynomial.  A polynomial of the form $x^n + 1$ is a multiple of $x + 1$ only when $n$ is odd.

So, we can try to get close by considering $\frac{x^{11} + 1}{x + 1}.$  Let's also multiply this by $x,$ so that we get a polynomial of degree 12.  Thus,
\begin{align*}
\frac{x^{12} - 1}{x - 1} \cdot \frac{x(x^{11} + 1)}{x + 1} &= \frac{x^{12} - 1}{x - 1} \cdot \frac{x^{12} + x}{x + 1} \\
&= \frac{x^{12} - 1}{x^2 - 1} \cdot (x^{12} + x) \\
&= (x^{10} + x^8 + x^6 + x^4 + x^2 + 1)(x^{12} + x) \\
&= x^{22} + x^{20} + x^{18} + x^{16} + x^{14} + x^{12} + x^{11} + x^9 + x^7 + x^5 + x^3 + x.
\end{align*}This is a multiple of $q(x)$ that's very close to $p(x).$  In fact, when we take the difference, we get
\begin{align*}
&p(x) - (x^{22} + x^{20} + x^{18} + x^{16} + x^{14} + x^{12} + x^{11} + x^9 + x^7 + x^5 + x^3 + x) \\
&\quad = -x^{11} + x^{10} - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1.
\end{align*}Now, if we add $q(x),$ we get
\begin{align*}
&p(x) + q(x) - (x^{22} + x^{20} + x^{18} + x^{16} + x^{14} + x^{12} + x^{11} + x^9 + x^7 + x^5 + x^3 + x) \\
&\quad = 2x^{10} + 2x^8 + 2x^6 + 2x^4 + 2x^2 + 2.
\end{align*}We can also write this as
\begin{align*}
&p(x) - (x^{22} + x^{20} + x^{18} + x^{16} + x^{14} + x^{12} + x^{11} + x^9 + x^7 + x^5 + x^3 + x - q(x)) \\
&\quad = 2x^{10} + 2x^8 + 2x^6 + 2x^4 + 2x^2 + 2.
\end{align*}So, we took $p(x),$ subtracted
\[x^{22} + x^{20} + x^{18} + x^{16} + x^{14} + x^{12} + x^{11} + x^9 + x^7 + x^5 + x^3 + x - q(x),\]which we know is a multiple of $q(x),$ and ended up with $2x^{10} + 2x^8 + 2x^6 + 2x^4 + 2x^2 + 2.$  Since the degree of this polynomial is less than the degree of $q(x),$ this is our remainder.
기하급수 공식에서,
\p(x) = 1 + x^2 + x^4 + x^6 + \점 + x^{22} = \frac{x^{24} - 1}{x^2 - 1}.\]도 마찬가지입니다,
\q(x) = 1 + x + x^2 + x^3 + \dots + x^{11} = \frac{x^{12} - 1}{x - 1}.\]언뜻 보기에 $p(x)$를 $q(x)$의 배수로 쓸 수 있는 것처럼 보일 수 있습니다:
\[\frac{x^{24} - 1}{x^2 - 1} = \frac{x^{12} - 1}{x - 1} \cdot \frac{x^{12} + 1}{x + 1}.\]불행히도, $\frac{x^{12} + 1}{x + 1}$는 다항식이 아닙니다. x^n + 1$ 형식의 다항식은 $n$이 홀수일 때만 $x + 1$의 배수입니다.

따라서 $\frac{x^{11} + 1}{x + 1}.$ 여기에 $x,$를 곱하면 차수 12의 다항식을 얻을 수 있습니다. 따라서
\begin{align*}
\frac{x^{12} - 1}{x - 1} \cdot \frac{x(x^{11} + 1)}{x + 1} &= \frac{x^{12} - 1}{x - 1} \cdot \frac{x^{12} + x}{x + 1} \\
&= \frac{x^{12} - 1}{x^2 - 1} \cdot (x^{12} + x) \\
&= (x^{10} + x^8 + x^6 + x^4 + x^2 + 1)(x^{12} + x) \\
&= x^{22} + x^{20} + x^{18} + x^{16} + x^{14} + x^{12} + x^{11} + x^9 + x^7 + x^5 + x^3 + x.
\end{align*}이것은 $q(x)$의 배수로 $p(x)$에 매우 가깝습니다. 실제로 차이를 취하면 다음과 같습니다.
\begin{align*}
&p(x) - (x^{22} + x^{20} + x^{18} + x^{16} + x^{14} + x^{12} + x^{11} + x^9 + x^7 + x^5 + x^3 + x) \\
&\쿼드 = -x^{11} + x^{10} - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1.
\end{align*}이제, $q(x)$를 더하면 다음과 같이 됩니다.
\begin{align*}
&p(x) + q(x) - (x^{22} + x^{20} + x^{18} + x^{16} + x^{14} + x^{12} + x^{11} + x^9 + x^7 + x^5 + x^3 + x) \\
&\쿼드 = 2x^{10} + 2x^8 + 2x^6 + 2x^4 + 2x^2 + 2.
\end{align*}이것을 다음과 같이 쓸 수도 있습니다.
\begin{align*}
&p(x) - (x^{22} + x^{20} + x^{18} + x^{16} + x^{14} + x^{12} + x^{11} + x^9 + x^7 + x^5 + x^3 + x - q(x)) \\
&\쿼드 = 2x^{10} + 2x^8 + 2x^6 + 2x^4 + 2x^2 + 2.
\end{align*}따라서, $p(x)$에서 다음과 같이 뺍니다.
\[x^{22} + x^{20} + x^{18} + x^{16} + x^{14} + x^{12} + x^{11} + x^9 + x^7 + x^5 + x^3 + x - q(x),\]는 $q(x)$의 배수이며, 결국 $2x^{10}이 됩니다. + 2x^8 + 2x^6 + 2x^4 + 2x^2 + 2.$ 이 다항식의 차수가 $q(x)$ 의 차수보다 작으므로 이것이 우리의 나머지 부분입니다.