다항식의 연산

식의 형태에서 a 와 b 를 찾는다.

이 식에서 $a=x$ 이고 $b=-2$ 이다.

이항정리를 이용한다.

$${(a+b)}^{n}=\left(\begin{array}{c}n \\ 0\end{array}\right){a}^{n}+\left(\begin{array}{c}n \\ 1\end{array}\right){a}^{n-1}{b}^{1}+\left(\begin{array}{c}n \\ 2\end{array}\right){a}^{n-2}{b}^{2}+...+\left(\begin{array}{c}n \\ r\end{array}\right){a}^{n-r}{b}^{r}+...+\left(\begin{array}{c}n \\ n\end{array}\right){b}^{n}$$

$a=x,$ $b=-2,$ $n=5$ 를 대입한다.

$${(x-2)}^{5}=\left(\begin{array}{c}5 \\ 0\end{array}\right){x}^{5}+\left(\begin{array}{c}5 \\ 1\end{array}\right){x}^{5-1}{(-2)}^{1}+\left(\begin{array}{c}5 \\ 2\end{array}\right){x}^{5-2}{(-2)}^{2}+\left(\begin{array}{c}5 \\ 3\end{array}\right){x}^{5-3}{(-2)}^{3}+\left(\begin{array}{c}5 \\ 4\end{array}\right){x}^{5-4}{(-2)}^{4}+\left(\begin{array}{c}5 \\ 5\end{array}\right){(-2)}^{5}$$

지수를 간단히 하고 계수의 값을 구한다. $\left(\begin{array}{c}n \\ 1\end{array}\right)=n,$ $\left(\begin{array}{c}n \\ n\end{array}\right)=1,$ $\left(\begin{array}{c}n \\ 0\end{array}\right)=1$ 임을 기억하자.

$$\begin{array}{c} \\ \\ {(x-2)}^{5}=\left(\begin{array}{c}5 \\ 0\end{array}\right){x}^{5}+\left(\begin{array}{c}5 \\ 1\end{array}\right){x}^{4}{(-2)}^{}+\left(\begin{array}{c}5 \\ 2\end{array}\right){x}^{3}{(-2)}^{2}+\left(\begin{array}{c}5 \\ 3\end{array}\right){x}^{2}{(-2)}^{3}+\left(\begin{array}{c}5 \\ 4\end{array}\right)x{(-2)}^{4}+\left(\begin{array}{c}5 \\ 5\end{array}\right){(-2)}^{5} \\ {(x-2)}^{5}=1{x}^{5}+5(-2){x}^{4}+\frac{5!}{2!\cdot 3!}{(-2)}^{2}{x}^{3}+\frac{5!}{3!\cdot 2!}{(-2)}^{3}{x}^{2}+\frac{5!}{4!\cdot 1!}{(-2)}^{4}x+1{(-2)}^{5} \\ {(x-2)}^{5}={x}^{5}+5(-2){x}^{4}+10\cdot 4\cdot {x}^{3}+10(-8){x}^{2}+5\cdot 16\cdot x+1(-32) \\ {(x-2)}^{5}={x}^{5}-10{x}^{4}+40{x}^{3}-80{x}^{2}+80x-32\end{array}$$