다항식 전체 문제

ⓐ We will use the definition of a binomial coefficient, $\left(\begin{array}{c}n \\ r\end{array}\right)=\frac{n!}{r!(n-r)!}.$

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ⓐ 6 ⓑ 1 ⓒ 1 ⓓ 35

ⓐ 2 ⓑ 1 ⓒ 1 ⓓ 6

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

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

${x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}$ $+5x{y}^{4}+{y}^{5}$

${m}^{6}+6{m}^{5}n+15{m}^{4}{n}^{2}+20{m}^{3}{n}^{3}$ $+15{m}^{2}{n}^{4}+6m{n}^{5}+{n}^{6}$

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

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

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

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

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

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