다항식의 연산

We identify the a and b of the pattern.

In our pattern, $a=p$ and $b=q.$

We use the Binomial Theorem.

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

Substitute in the values $a=p,$ $b=q$ and $n=4.$

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

Simplify the exponents.

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

Evaluate the coefficients. Remember, $\left(\begin{array}{c}n \\ 1\end{array}\right)=n,$ $\left(\begin{array}{c}n \\ n\end{array}\right)=1,$ $\left(\begin{array}{c}n \\ 0\end{array}\right)=1.$

$$\begin{array}{c}{(p+q)}^{4}=1{p}^{4}+4{p}^{3}{q}^{1}+\frac{4!}{2!(2)!}{p}^{2}{q}^{2}+\frac{4!}{3!(4-3)!}{p}^{1}{q}^{3}+1{q}^{4} \\ {(p+q)}^{4}={p}^{4}+4{p}^{3}{q}^{}+6{p}^{2}{q}^{2}+4{p}^{}{q}^{3}+{q}^{4}\end{array}$$