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Binomial Theorem

Use Pascal’s Triangle to expand a binomial

Use Pascal’s Triangle to Expand a Binomial

In our previous work, we have squared binomials either by using FOIL or by using the Binomial Squares Pattern. We can also say that we expanded \({(a+b)}^{2}.\)

\[{(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\]

To expand \({(a+b)}^{3},\) we recognize that this is \({(a+b)}^{2}(a+b)\) and multiply.

\[\begin{array}{l}{(a+b)}^{3} \\ {(a+b)}^{2}(a+b) \\ ({a}^{2}+2ab+{b}^{2})(a+b) \\ {a}^{3}+2{a}^{2}b+a{b}^{2}+{a}^{2}b+2a{b}^{2}+{b}^{3} \\ {a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3} \\ {(a+b)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}\end{array}\]

To find a method that is less tedious that will work for higher expansions like \({(a+b)}^{7},\) we again look for patterns in some expansions.

	Number of terms	First term	Last term
\({(a+b)}^{1}=a+b\)	2	\({a}^{1}\)	\({b}^{1}\)
\({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\)	3	\({a}^{2}\)	\({b}^{2}\)
\({(a+b)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}\)	4	\({a}^{3}\)	\({b}^{3}\)
\({(a+b)}^{4}={a}^{4}+4{a}^{3}b+6{a}^{2}{b}^{2}+4a{b}^{3}+{b}^{4}\)	5	\({a}^{4}\)	\({b}^{4}\)
\({(a+b)}^{5}={a}^{5}+5{a}^{4}b+10{a}^{3}{b}^{2}+10{a}^{2}{b}^{3}+5a{b}^{4}+{b}^{5}\)	6	\({a}^{5}\)	\({b}^{5}\)
\({(a+b)}^{n}\)	\(+1\)	\({a}^{n}\)	\({b}^{n}\)

Notice the first and last terms show only one variable. Recall that \({a}^{0}=1,\) so we could rewrite the first and last terms to include both variables. For example, we could expand \({(a+b)}^{3}\) to show each term with both variables.

Generally, we don’t show the zero exponents, just as we usually write x rather than 1x.

Let’s look at an example to highlight the last three patterns.

From the patterns we identified, we see the variables in the expansion of \({(a+b)}^{n},\) would be

\[{(a+b)}^{n}={a}^{n}+___{a}^{n-1}{b}^{1}+___{a}^{n-2}{b}^{2}+...+___{a}^{1}{b}^{n-1}+{b}^{n}.\]

Example

Try it.

Use Pascal’s Triangle to expand \({(x+y)}^{6}.\)

Solution

We know the variables for this expansion will follow the pattern we identified. The nonzero exponents of x will start at six and decrease to one. The nonzero exponents of y will start at one and increase to six. The sum of the exponents in each term will be six. In our pattern, \(a=x\) and \(b=y.\)

\[\begin{array}{l}{(a+b)}^{n}={a}^{n}+___{a}^{n-1}{b}^{1}+___{a}^{n-2}{b}^{2}+...+___{a}^{1}{b}^{n-1}+{b}^{n} \\ {(x+y)}^{6}={x}^{6}+___{x}^{5}{y}^{1}+___{x}^{4}{y}^{2}+___{x}^{3}{y}^{3}+___{x}^{2}{y}^{4}+___{x}^{1}{y}^{5}+{y}^{6}\end{array}\]

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Evaluate a Binomial Coefficient

While Pascal’s Triangle is one method to expand a binomial, we will also look at another method. Before we get to that, we need to introduce some more factorial notation. This notation is not only used to expand binomials, but also in the study and use of probability.

To find the coefficients of the terms of expanded binomials, we will need to be able to evaluate the notation \((\begin{array}{l}n \\ r\end{array})\) which is called a binomial coefficient. We read \((\begin{array}{l}n \\ r\end{array})\) as “n choose r” or “n taken r at a time”.

Example

Try it.

Evaluate: ⓐ \((\begin{array}{l}5 \\ 1\end{array})\) ⓑ \((\begin{array}{l}7 \\ 7\end{array})\) ⓒ \((\begin{array}{l}4 \\ 0\end{array})\) ⓓ \((\begin{array}{l}8 \\ 5\end{array}).\)

Solution

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

	\(\ (\begin{array}{l}5 \\ 1\end{array})\)
Use the definition, \((\begin{array}{l}n \\ r\end{array})=\frac{n!}{r!(n-r)!},\) where \(n=5,r=1.\)	\(\ \frac{5!}{1!(5-1)!}\)
Simplify.	\(\ \frac{5!}{1!(4)!}\)
Rewrite \(5!\ \text{as}\ 5\cdot 4!\)	\(\ \frac{5\cdot 4!}{1!\cdot 4!}\)
Simplify, by removing common factors.	\(\ \frac{5\cdot 4!}{1\cdot 4!}\)
Simplify.	\(\ 5\)
	\(\ (\begin{array}{l}5 \\ 1\end{array})=5\)

ⓑ

	\(\ (\begin{array}{l}7 \\ 7\end{array})\)
Use the definition, \((\begin{array}{l}n \\ r\end{array})=\frac{n!}{r!(n-r)!},\) where \(n=7,r=7.\)	\(\ \frac{7!}{7!(7-7)!}\)
Simplify.	\(\ \frac{7!}{7!(0)!}\)
Simplify. Remember \(0!=1.\)	\(\ 1\)
	\(\ (\begin{array}{l}7 \\ 7\end{array})=1\)

ⓒ

	\(\ (\begin{array}{l}4 \\ 0\end{array})\)
Use the definition, \((\begin{array}{l}n \\ r\end{array})=\frac{n!}{r!(n-r)!},\) where \(n=4,r=0.\)	\(\ \frac{4!}{0!(4-0)!}\)
Simplify.	\(\ \frac{4!}{0!(4)!}\)
Simplify.	\(\ 1\)
	\(\ (\begin{array}{l}4 \\ 0\end{array})=1\)

ⓓ

	\(\ (\begin{array}{l}8 \\ 5\end{array})\)
Use the definition, \((\begin{array}{l}n \\ r\end{array})=\frac{n!}{r!(n-r)!},\) where \(n=8,r=5.\)	\(\ \frac{8!}{5!(8-5)!}\)
Simplify.	\(\ \frac{8!}{5!(3)!}\)
Rewrite \(8!\ \text{as}\ 8\cdot 7\cdot 6\cdot 5!\) and remove common factors.	\(\ \frac{8\cdot 7\cdot 6\cdot 5!}{5!\cdot 3\cdot 2\cdot 1}\)
Simplify.	\(\ 56\)
	\(\ (\begin{array}{l}8 \\ 5\end{array})=56\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Use the Binomial Theorem to Expand a Binomial

We are now ready to use the alternate method of expanding binomials. The Binomial Theorem uses the same pattern for the variables, but uses the binomial coefficient for the coefficient of each term.

Example

Try it.

Use the Binomial Theorem to expand \({(p+q)}^{4}.\)

Solution

We identify the a and b of the pattern.

In our pattern, \(a=p\) and \(b=q.\)

We use the Binomial Theorem.

\[{(a+b)}^{n}=(\begin{array}{l}n \\ 0\end{array}){a}^{n}+(\begin{array}{l}n \\ 1\end{array}){a}^{n-1}{b}^{1}+(\begin{array}{l}n \\ 2\end{array}){a}^{n-2}{b}^{2}+...+(\begin{array}{l}n \\ r\end{array}){a}^{n-r}{b}^{r}+...+(\begin{array}{l}n \\ n\end{array}){b}^{n}\]

Substitute in the values \(a=p,\)\(b=q\) and \(n=4.\)

\[{(p+q)}^{4}=(\begin{array}{l}4 \\ 0\end{array}){p}^{4}+(\begin{array}{l}4 \\ 1\end{array}){p}^{4-1}{q}^{1}+(\begin{array}{l}4 \\ 2\end{array}){p}^{4-2}{q}^{2}+(\begin{array}{l}4 \\ 3\end{array}){p}^{4-3}{q}^{3}+(\begin{array}{l}4 \\ 4\end{array}){q}^{4}\]

Simplify the exponents.

\[{(p+q)}^{4}=(\begin{array}{l}4 \\ 0\end{array}){p}^{4}+(\begin{array}{l}4 \\ 1\end{array}){p}^{3}{q}^{}+(\begin{array}{l}4 \\ 2\end{array}){p}^{2}{q}^{2}+(\begin{array}{l}4 \\ 3\end{array}){p}^{}{q}^{3}+(\begin{array}{l}4 \\ 4\end{array}){q}^{4}\]

Evaluate the coefficients. Remember, \((\begin{array}{l}n \\ 1\end{array})=n,\)\((\begin{array}{l}n \\ n\end{array})=1,\)\((\begin{array}{l}n \\ 0\end{array})=1.\)

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

Notice that when we expanded \({(p+q)}^{4}\) in the last example, using the Binomial Theorem, we got the same coefficients we would get from using Pascal’s Triangle.

The next example, the binomial is a difference. When the binomial is a difference, we must be careful in identifying the values we will use in the pattern.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

Patterns in the expansion of \({(a+b)}^{n}\)
- The number of terms is \(n+1.\)
- The first term is \({a}^{n}\) and the last term is \({b}^{n}.\)
- The exponents on a decrease by one on each term going left to right.
- The exponents on b increase by one on each term going left to right.
- The sum of the exponents on any term is n.
Pascal’s Triangle
Binomial Coefficient \((\begin{array}{l}n \\ r\end{array})\) : A binomial coefficient \((\begin{array}{l}n \\ r\end{array}),\) where r and n are integers with \(0\le r\le n,\) is defined as
\[(\begin{array}{l}n \\ r\end{array})=\frac{n!}{r!(n-r)!}\]
We read \((\begin{array}{l}n \\ r\end{array})\) as “n choose r” or “n taken r at a time”.
Properties of Binomial Coefficients
\(\begin{array}{lllllllll}(\begin{array}{l}n \\ 1\end{array})=n & & & & (\begin{array}{l}n \\ n\end{array})=1 & & & & (\begin{array}{l}n \\ 0\end{array})=1\end{array}\)
Binomial Theorem: For any real numbers a, b, and positive integer n,
\[{(a+b)}^{n}=(\begin{array}{l}n \\ 0\end{array}){a}^{n}+(\begin{array}{l}n \\ 1\end{array}){a}^{n-1}{b}^{1}+(\begin{array}{l}n \\ 2\end{array}){a}^{n-2}{b}^{2}+...+(\begin{array}{l}n \\ r\end{array}){a}^{n-r}{b}^{r}+...+(\begin{array}{l}n \\ n\end{array}){b}^{n}\]

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Simplify: \(\frac{7\cdot 6\cdot 5\cdot 4}{4\cdot 3\cdot 2\cdot 1}.\)
If you missed this problem, review .

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35
Expand: \({(3x+5)}^{2}.\)
If you missed this problem, review .

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\(9{x}^{2}+30x+25\)
Expand: \({(x-y)}^{2}.\)
If you missed this problem, review .

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\({x}^{2}-2xy+{y}^{2}\)
Use Pascal’s Triangle to expand \({(x+y)}^{6}.\)

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We know the variables for this expansion will follow the pattern we identified. The nonzero exponents of x will start at six and decrease to one. The nonzero exponents of y will start at one and increase to six. The sum of the exponents in each term will be six. In our pattern, \(a=x\) and \(b=y.\)
\[\begin{array}{l}{(a+b)}^{n}={a}^{n}+___{a}^{n-1}{b}^{1}+___{a}^{n-2}{b}^{2}+...+___{a}^{1}{b}^{n-1}+{b}^{n} \\ {(x+y)}^{6}={x}^{6}+___{x}^{5}{y}^{1}+___{x}^{4}{y}^{2}+___{x}^{3}{y}^{3}+___{x}^{2}{y}^{4}+___{x}^{1}{y}^{5}+{y}^{6}\end{array}\]
Use Pascal’s Triangle to expand \({(x+y)}^{5}.\)

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\({x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}\)
\(+5x{y}^{4}+{y}^{5}\)
Use Pascal’s Triangle to expand \({(p+q)}^{7}.\)

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\({p}^{7}+7{p}^{6}q+21{p}^{5}{q}^{2}+35{p}^{4}{q}^{3}\)
\(+35{p}^{3}{q}^{4}+21{p}^{2}{q}^{5}+7p{q}^{6}+{q}^{7}\)
Use Pascal’s Triangle to expand \({(x+3)}^{5}.\)

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We identify the a and b of the pattern.
In our pattern, \(a=x\) and \(b=3.\)
We know the variables for this expansion will follow the pattern we identified. The sum of the exponents in each term will be five.
\[\begin{array}{l}{(a+b)}^{n}={a}^{n}+___{a}^{n-1}{b}^{1}+___{a}^{n-2}{b}^{2}+...+___{a}^{1}{b}^{n-1}+{b}^{n} \\ {(x+3)}^{5}={x}^{5}+___{x}^{4}\cdot {3}^{1}+___{x}^{3}\cdot {3}^{2}+___{x}^{2}\cdot {3}^{3}+___{x}^{1}\cdot {3}^{4}+{3}^{5}\end{array}\]
Use Pascal’s Triangle to expand \({(x+2)}^{4}.\)

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\({x}^{4}+8{x}^{3}+24{x}^{2}+32x+16\)
Use Pascal’s Triangle to expand \({(x+1)}^{6}.\)

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\({x}^{6}+6{x}^{5}+15{x}^{4}+20{x}^{3}+15{x}^{2}\)
\(+6x+1\)
Use Pascal’s Triangle to expand \({(3x-2)}^{4}.\)

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We identify the a and b of the pattern.
In our pattern, \(a=3x\) and \(b=-2.\)

\[\begin{array}{l}{(a+b)}^{n}={a}^{n}+___{a}^{n-1}{b}^{1}+___{a}^{n-2}{b}^{2}+...+___{a}^{1}{b}^{n-1}+{b}^{n} \\ {(3x-2)}^{4}=1\cdot {(3x)}^{4}+4{(3x)}^{3}{(-2)}^{1}+6{(3x)}^{2}{(-2)}^{2}+4{(3x)}^{1}{(-2)}^{3}+1\cdot {(-2)}^{4} \\ {(3x-2)}^{4}=81{x}^{4}+4(27{x}^{3})(-2)+6(9{x}^{2})(4)+4(3x)(-8)+1\cdot 16 \\ {(3x-2)}^{4}=81{x}^{4}-216{x}^{3}+216{x}^{2}-96x+16\end{array}\]
Use Pascal’s Triangle to expand \({(2x-3)}^{4}.\)

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\(16{x}^{4}-96{x}^{3}+216{x}^{2}-216x+81\)
Use Pascal’s Triangle to expand \({(2x-1)}^{6}.\)

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\(64{x}^{6}-192{x}^{5}+240{x}^{4}-160{x}^{3}\)
\(+60{x}^{2}-12x+1\)

Evaluate: ⓐ \((\begin{array}{l}5 \\ 1\end{array})\) ⓑ \((\begin{array}{l}7 \\ 7\end{array})\) ⓒ \((\begin{array}{l}4 \\ 0\end{array})\) ⓓ \((\begin{array}{l}8 \\ 5\end{array}).\)

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

	\(\ (\begin{array}{l}5 \\ 1\end{array})\)
Use the definition, \((\begin{array}{l}n \\ r\end{array})=\frac{n!}{r!(n-r)!},\) where \(n=5,r=1.\)	\(\ \frac{5!}{1!(5-1)!}\)
Simplify.	\(\ \frac{5!}{1!(4)!}\)
Rewrite \(5!\ \text{as}\ 5\cdot 4!\)	\(\ \frac{5\cdot 4!}{1!\cdot 4!}\)
Simplify, by removing common factors.	\(\ \frac{5\cdot 4!}{1\cdot 4!}\)
Simplify.	\(\ 5\)
	\(\ (\begin{array}{l}5 \\ 1\end{array})=5\)

ⓑ

	\(\ (\begin{array}{l}7 \\ 7\end{array})\)
Use the definition, \((\begin{array}{l}n \\ r\end{array})=\frac{n!}{r!(n-r)!},\) where \(n=7,r=7.\)	\(\ \frac{7!}{7!(7-7)!}\)
Simplify.	\(\ \frac{7!}{7!(0)!}\)
Simplify. Remember \(0!=1.\)	\(\ 1\)
	\(\ (\begin{array}{l}7 \\ 7\end{array})=1\)

ⓒ

	\(\ (\begin{array}{l}4 \\ 0\end{array})\)
Use the definition, \((\begin{array}{l}n \\ r\end{array})=\frac{n!}{r!(n-r)!},\) where \(n=4,r=0.\)	\(\ \frac{4!}{0!(4-0)!}\)
Simplify.	\(\ \frac{4!}{0!(4)!}\)
Simplify.	\(\ 1\)
	\(\ (\begin{array}{l}4 \\ 0\end{array})=1\)

ⓓ

	\(\ (\begin{array}{l}8 \\ 5\end{array})\)
Use the definition, \((\begin{array}{l}n \\ r\end{array})=\frac{n!}{r!(n-r)!},\) where \(n=8,r=5.\)	\(\ \frac{8!}{5!(8-5)!}\)
Simplify.	\(\ \frac{8!}{5!(3)!}\)
Rewrite \(8!\ \text{as}\ 8\cdot 7\cdot 6\cdot 5!\) and remove common factors.	\(\ \frac{8\cdot 7\cdot 6\cdot 5!}{5!\cdot 3\cdot 2\cdot 1}\)
Simplify.	\(\ 56\)
	\(\ (\begin{array}{l}8 \\ 5\end{array})=56\)

Evaluate each binomial coefficient:
ⓐ \((\begin{array}{l}6 \\ 1\end{array})\) ⓑ \((\begin{array}{l}8 \\ 8\end{array})\) ⓒ \((\begin{array}{l}5 \\ 0\end{array})\) ⓓ \((\begin{array}{l}7 \\ 3\end{array}).\)

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ⓐ 6 ⓑ 1 ⓒ 1 ⓓ 35
Evaluate each binomial coefficient:
ⓐ \((\begin{array}{l}2 \\ 1\end{array})\) ⓑ \((\begin{array}{l}11 \\ 11\end{array})\) ⓒ \((\begin{array}{l}9 \\ 0\end{array})\) ⓓ \((\begin{array}{l}6 \\ 5\end{array}).\)

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ⓐ 2 ⓑ 1 ⓒ 1 ⓓ 6
Use the Binomial Theorem to expand \({(p+q)}^{4}.\)

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We identify the a and b of the pattern.
In our pattern, \(a=p\) and \(b=q.\)
We use the Binomial Theorem.
\[{(a+b)}^{n}=(\begin{array}{l}n \\ 0\end{array}){a}^{n}+(\begin{array}{l}n \\ 1\end{array}){a}^{n-1}{b}^{1}+(\begin{array}{l}n \\ 2\end{array}){a}^{n-2}{b}^{2}+...+(\begin{array}{l}n \\ r\end{array}){a}^{n-r}{b}^{r}+...+(\begin{array}{l}n \\ n\end{array}){b}^{n}\]
Substitute in the values \(a=p,\)\(b=q\) and \(n=4.\)
\[{(p+q)}^{4}=(\begin{array}{l}4 \\ 0\end{array}){p}^{4}+(\begin{array}{l}4 \\ 1\end{array}){p}^{4-1}{q}^{1}+(\begin{array}{l}4 \\ 2\end{array}){p}^{4-2}{q}^{2}+(\begin{array}{l}4 \\ 3\end{array}){p}^{4-3}{q}^{3}+(\begin{array}{l}4 \\ 4\end{array}){q}^{4}\]
Simplify the exponents.
\[{(p+q)}^{4}=(\begin{array}{l}4 \\ 0\end{array}){p}^{4}+(\begin{array}{l}4 \\ 1\end{array}){p}^{3}{q}^{}+(\begin{array}{l}4 \\ 2\end{array}){p}^{2}{q}^{2}+(\begin{array}{l}4 \\ 3\end{array}){p}^{}{q}^{3}+(\begin{array}{l}4 \\ 4\end{array}){q}^{4}\]
Evaluate the coefficients. Remember, \((\begin{array}{l}n \\ 1\end{array})=n,\)\((\begin{array}{l}n \\ n\end{array})=1,\)\((\begin{array}{l}n \\ 0\end{array})=1.\)
\[\begin{array}{l}{(p+q)}^{4}=1{p}^{4}+4{p}^{3}{q}^{1}+\frac{4!}{2!(2)!}{p}^{2}{q}^{2}+\frac{4!}{3!(4-3)!}{p}^{1}{q}^{3}+1{q}^{4} \\ {(p+q)}^{4}={p}^{4}+4{p}^{3}{q}^{}+6{p}^{2}{q}^{2}+4{p}^{}{q}^{3}+{q}^{4}\end{array}\]
Use the Binomial Theorem to expand \({(x+y)}^{5}.\)

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\({x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}\)
\(+5x{y}^{4}+{y}^{5}\)
Use the Binomial Theorem to expand \({(m+n)}^{6}.\)

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\({m}^{6}+6{m}^{5}n+15{m}^{4}{n}^{2}+20{m}^{3}{n}^{3}\)
\(+15{m}^{2}{n}^{4}+6m{n}^{5}+{n}^{6}\)
Use the Binomial Theorem to expand \({(x-2)}^{5}.\)

Αποκάλυψέ την.

We identify the a and b of the pattern.
In our pattern, \(a=x\) and \(b=-2.\)
We use the Binomial Theorem.
\[{(a+b)}^{n}=(\begin{array}{l}n \\ 0\end{array}){a}^{n}+(\begin{array}{l}n \\ 1\end{array}){a}^{n-1}{b}^{1}+(\begin{array}{l}n \\ 2\end{array}){a}^{n-2}{b}^{2}+...+(\begin{array}{l}n \\ r\end{array}){a}^{n-r}{b}^{r}+...+(\begin{array}{l}n \\ n\end{array}){b}^{n}\]
Substitute in the values \(a=x,\)\(b=-2,\) and \(n=5.\)
\[{(x-2)}^{5}=(\begin{array}{l}5 \\ 0\end{array}){x}^{5}+(\begin{array}{l}5 \\ 1\end{array}){x}^{5-1}{(-2)}^{1}+(\begin{array}{l}5 \\ 2\end{array}){x}^{5-2}{(-2)}^{2}+(\begin{array}{l}5 \\ 3\end{array}){x}^{5-3}{(-2)}^{3}+(\begin{array}{l}5 \\ 4\end{array}){x}^{5-4}{(-2)}^{4}+(\begin{array}{l}5 \\ 5\end{array}){(-2)}^{5}\]
Simplify the exponents and evaluate the coefficients. Remember,\((\begin{array}{l}n \\ 1\end{array})=n,\)\((\begin{array}{l}n \\ n\end{array})=1,\)\((\begin{array}{l}n \\ 0\end{array})=1.\)
\[\begin{array}{l} \\ \\ {(x-2)}^{5}=(\begin{array}{l}5 \\ 0\end{array}){x}^{5}+(\begin{array}{l}5 \\ 1\end{array}){x}^{4}{(-2)}^{}+(\begin{array}{l}5 \\ 2\end{array}){x}^{3}{(-2)}^{2}+(\begin{array}{l}5 \\ 3\end{array}){x}^{2}{(-2)}^{3}+(\begin{array}{l}5 \\ 4\end{array})x{(-2)}^{4}+(\begin{array}{l}5 \\ 5\end{array}){(-2)}^{5} \\ {(x-2)}^{5}=1{x}^{5}+5(-2){x}^{4}+\frac{5!}{2!\cdot 3!}{(-2)}^{2}{x}^{3}+\frac{5!}{3!\cdot 2!}{(-2)}^{3}{x}^{2}+\frac{5!}{4!\cdot 1!}{(-2)}^{4}x+1{(-2)}^{5} \\ {(x-2)}^{5}={x}^{5}+5(-2){x}^{4}+10\cdot 4\cdot {x}^{3}+10(-8){x}^{2}+5\cdot 16\cdot x+1(-32) \\ {(x-2)}^{5}={x}^{5}-10{x}^{4}+40{x}^{3}-80{x}^{2}+80x-32\end{array}\]
Use the Binomial Theorem to expand \({(x-3)}^{5}.\)

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\({x}^{5}-15{x}^{4}+90{x}^{3}-270{x}^{2}\)
\(+405x-243\)
Use the Binomial Theorem to expand \({(y-1)}^{6}.\)

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\({y}^{6}-6{y}^{5}+15{y}^{4}-20{y}^{3}+15{y}^{2}\)
\(-6y+1\)
Use the Binomial Theorem to expand \({(2x-3y)}^{4}.\)

Αποκάλυψέ την.

We identify the a and b of the pattern.
In our pattern, \(a=2x\) and \(b=-3y.\)
We use the Binomial Theorem.
\[{(a+b)}^{n}=(\begin{array}{l}n \\ 0\end{array}){a}^{n}+(\begin{array}{l}n \\ 1\end{array}){a}^{n-1}{b}^{1}+(\begin{array}{l}n \\ 2\end{array}){a}^{n-2}{b}^{2}+...+(\begin{array}{l}n \\ r\end{array}){a}^{n-r}{b}^{r}+...+(\begin{array}{l}n \\ n\end{array}){b}^{n}\]
Substitute in the values \(a=2x,\)\(b=-3y\) and \(n=4.\)
\[{(2x-3y)}^{4}=(\begin{array}{l}4 \\ 0\end{array}){(2x)}^{4}+(\begin{array}{l}4 \\ 1\end{array}){(2x)}^{4-1}{(-3y)}^{1}+(\begin{array}{l}4 \\ 2\end{array}){(2x)}^{4-2}{(-3y)}^{2}+(\begin{array}{l}4 \\ 3\end{array}){(2x)}^{4-3}{(-3y)}^{3}+(\begin{array}{l}4 \\ 4\end{array}){(-3y)}^{4}\]
Simplify the exponents.
\[{(2x-3y)}^{4}=(\begin{array}{l}4 \\ 0\end{array}){(2x)}^{4}+(\begin{array}{l}4 \\ 1\end{array}){(2x)}^{3}{(-3y)}^{1}+(\begin{array}{l}4 \\ 2\end{array}){(2x)}^{2}{(-3y)}^{2}+(\begin{array}{l}4 \\ 3\end{array}){(2x)}^{1}{(-3y)}^{3}+(\begin{array}{l}4 \\ 4\end{array}){(-3y)}^{4}\]
Evaluate the coefficients. Remember, \((\begin{array}{l}n \\ 1\end{array})=n,\ (\begin{array}{l}n \\ n\end{array})=1,\ (\begin{array}{l}n \\ 0\end{array})=1.\)
\({(2x-3y)}^{4}=1{(2x)}^{4}+4{(2x)}^{3}{(-3y)}^{1}+\frac{4!}{2!(2)!}{(2x)}^{2}{(-3y)}^{2}+\frac{4!}{3!(4-3)!}{(2x)}^{1}{(-3y)}^{3}+1{(-3y)}^{4}\)
\({(2x-3y)}^{4}=16{x}^{4}+4\cdot 8{x}^{3}(-3y)+6(4{x}^{2})(9{y}^{2})+4(2x)(-27{y}^{3})+81{y}^{4}\)
\({(2x-3y)}^{4}=16{x}^{4}-96{x}^{3}y+216{x}^{2}{y}^{2}-216x{y}^{3}+81{y}^{4}\)
Use the Binomial Theorem to expand \({(3x-2y)}^{5}.\)

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\(243{x}^{5}-810{x}^{4}y+1080{x}^{3}{y}^{2}\)
\(-720{x}^{2}{y}^{3}+240x{y}^{4}-32{y}^{5}\)
Use the Binomial Theorem to expand \({(4x-3y)}^{4}.\)

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\(256{x}^{4}-768{x}^{3}y+864{x}^{2}{y}^{2}\)
\(-432x{y}^{3}+81{y}^{4}\)

Find the fourth term of \({(x+y)}^{7}.\)

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In our pattern, \(n=7,\ a=x\) and \(b=y.\)
We are looking for the fourth term. \(\ \begin{array}{lll}\text{Since}\ r+1 & = & 4, \\ \text{then}\ r & = & 3.\end{array}\)
Write the formula.
Substitute in the values, \(n=7,\)\(r=3,\)\(a=x,\) and \(b=y.\)

Simplify.
Simplify.

Find the third term of \({(x+y)}^{6}.\)

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\(15{x}^{4}{y}^{2}\)
Find the fifth term of \({(a+b)}^{8}.\)

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\(70{a}^{4}{b}^{4}\)

Find the coefficient of the \({x}^{6}\) term of \({(x+3)}^{9}.\)

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In our pattern, then \(n=9,\)\(a=x,\) and \(b=3.\)
We are looking for the coefficient of the \({x}^{6}\) term. Since \(a=x,\) and \({x}^{9-r}={x}^{6},\) we know \(r=3.\)
Write the formula.
Substitute in the values, \(n=9,\)\(r=3,\)\(a=x,\) and \(b=3.\)

Simplify.
Simplify.
Simplify.
	The coefficient of the \({x}^{6}\) term is 2268.

Find the coefficient of the \({x}^{5}\) term of \({(x+4)}^{8}.\)

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3,584
Find the coefficient of the \({x}^{4}\) term of \({(x+2)}^{7}.\)

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280
\({(x+y)}^{4}\)
\({(a+b)}^{8}\)

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\({a}^{8}+8{a}^{7}b+28{a}^{6}{b}^{2}+56{a}^{5}{b}^{3}\)
\(+70{a}^{4}{b}^{4}+56{a}^{3}{b}^{5}+28{a}^{2}{b}^{6}\)
\(+8a{b}^{7}+{b}^{8}\)
\({(m+n)}^{10}\)
\({(p+q)}^{9}\)

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\({p}^{9}+9{p}^{8}q+36{p}^{7}{q}^{2}+84{p}^{6}{q}^{3}\)
\(+126{p}^{5}{q}^{4}+126{p}^{4}{q}^{5}+84{p}^{3}{q}^{6}\)
\(+36{p}^{2}{q}^{7}+9p{q}^{8}+{q}^{9}\)
\({(x-y)}^{5}\)
\({(a-b)}^{6}\)

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\({a}^{6}-6{a}^{5}b+15{a}^{4}{b}^{2}-20{a}^{3}{b}^{3}\)
\(+15{a}^{2}{b}^{4}-6a{b}^{5}+{b}^{6}\)
\({(x+4)}^{4}\)
\({(x+5)}^{3}\)

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\({x}^{3}+15{x}^{2}+75x+125\)
\({(y+2)}^{5}\)
\({(y+1)}^{7}\)

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\({y}^{7}+7{y}^{6}+21{y}^{5}+35{y}^{4}+35{y}^{3}\)
\(+21{y}^{2}+7y+1\)

Symbols used here

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Binomial Theorem

Use Pascal’s Triangle to expand a binomial
Evaluate a binomial coefficient
Use the Binomial Theorem to expand a binomial
The number of terms is
The first term is
The exponents on
The exponents on
The sum of the exponents on any term is

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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