다항식의 연산

We identify the a and b of the pattern.

In our pattern, $a=x$ and $b=-2.$

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 $a=x,$ $b=-2,$ and $n=5.$

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

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

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

${x}^{5}-15{x}^{4}+90{x}^{3}-270{x}^{2}$ $+405x-243$

${y}^{6}-6{y}^{5}+15{y}^{4}-20{y}^{3}+15{y}^{2}$ $-6y+1$

We identify the a and b of the pattern.

In our pattern, $a=2x$ and $b=-3y.$

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 $a=2x,$ $b=-3y$ and $n=4.$

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

Simplify the exponents.

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

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

${(2x-3y)}^{4}=1{(2x)}^{4}+4{(2x)}^{3}{(-3y)}^{1}+\frac{4!}{2!(2)!}{(2x)}^{2}{(-3y)}^{2}+\frac{4!}{3!(4-3)!}{(2x)}^{1}{(-3y)}^{3}+1{(-3y)}^{4}$

${(2x-3y)}^{4}=16{x}^{4}+4\cdot 8{x}^{3}(-3y)+6(4{x}^{2})(9{y}^{2})+4(2x)(-27{y}^{3})+81{y}^{4}$

${(2x-3y)}^{4}=16{x}^{4}-96{x}^{3}y+216{x}^{2}{y}^{2}-216x{y}^{3}+81{y}^{4}$

$243{x}^{5}-810{x}^{4}y+1080{x}^{3}{y}^{2}$ $-720{x}^{2}{y}^{3}+240x{y}^{4}-32{y}^{5}$

$256{x}^{4}-768{x}^{3}y+864{x}^{2}{y}^{2}$ $-432x{y}^{3}+81{y}^{4}$

$15{x}^{4}{y}^{2}$

$70{a}^{4}{b}^{4}$

3,584

280

${a}^{8}+8{a}^{7}b+28{a}^{6}{b}^{2}+56{a}^{5}{b}^{3}$ $+70{a}^{4}{b}^{4}+56{a}^{3}{b}^{5}+28{a}^{2}{b}^{6}$ $+8a{b}^{7}+{b}^{8}$

${p}^{9}+9{p}^{8}q+36{p}^{7}{q}^{2}+84{p}^{6}{q}^{3}$ $+126{p}^{5}{q}^{4}+126{p}^{4}{q}^{5}+84{p}^{3}{q}^{6}$ $+36{p}^{2}{q}^{7}+9p{q}^{8}+{q}^{9}$

${a}^{6}-6{a}^{5}b+15{a}^{4}{b}^{2}-20{a}^{3}{b}^{3}$ $+15{a}^{2}{b}^{4}-6a{b}^{5}+{b}^{6}$

${x}^{3}+15{x}^{2}+75x+125$