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Studienführer > College Algebra

Use the Binomial Theorem

Learning Objectives

  • Identify binomial coefficients given the formula for a combination
  • Expand a binomial using the binomial theorem
  • Use the binomial to find a single term in a binomial
  Previously, we studied combinations. In the shortcut to finding [latex]{\left(x+y\right)}^{n}[/latex], we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation [latex]\left(\begin{array}{c}n\\ r\end{array}\right)[/latex] instead of [latex]C\left(n,r\right)[/latex], but it can be calculated in the same way. So

[latex]\left(\begin{array}{c}n\\ r\end{array}\right)=C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}[/latex]

The combination [latex]\left(\begin{array}{c}n\\ r\end{array}\right)[/latex] is called a binomial coefficient. An example of a binomial coefficient is [latex]\left(\begin{array}{c}5\\ 2\end{array}\right)=C\left(5,2\right)=10[/latex].

A General Note: Binomial Coefficients

If [latex]n[/latex] and [latex]r[/latex] are integers greater than or equal to 0 with [latex]n\ge r[/latex], then the binomial coefficient is

[latex]\left(\begin{array}{c}n\\ r\end{array}\right)=C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}

[/latex]

Q & A

Is a binomial coefficient always a whole number?

Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number.

Example: Finding Binomial Coefficients

Find each binomial coefficient.
  1. [latex]\left(\begin{array}{c}5\\ 3\end{array}\right)[/latex]
  2. [latex]\left(\begin{array}{c}9\\ 2\end{array}\right)[/latex]
  3. [latex]\left(\begin{array}{c}9\\ 7\end{array}\right)[/latex]

Answer: Use the formula to calculate each binomial coefficient. You can also use the [latex]{n}_{}{C}_{r}[/latex] function on your calculator.

[latex]\left(\begin{array}{c}n\\ r\end{array}\right)=C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}[/latex]

  1. [latex]\left(\begin{array}{c}5\\ 3\end{array}\right)=\frac{5!}{3!\left(5 - 3\right)!}=\frac{5\cdot 4\cdot 3!}{3!2!}=10[/latex]
  2. [latex]\left(\begin{array}{c}9\\ 2\end{array}\right)=\frac{9!}{2!\left(9 - 2\right)!}=\frac{9\cdot 8\cdot 7!}{2!7!}=36[/latex]
  3. [latex]\left(\begin{array}{c}9\\ 7\end{array}\right)=\frac{9!}{7!\left(9 - 7\right)!}=\frac{9\cdot 8\cdot 7!}{7!2!}=36[/latex]

Analysis of the Solution

Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations.

[latex]\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right)[/latex]

Try It

Find each binomial coefficient.
  1. [latex]\left(\begin{array}{c}7\\ 3\end{array}\right)[/latex]
  2. [latex]\left(\begin{array}{c}11\\ 4\end{array}\right)[/latex]

Answer:

  1. 35
  2. 330

When we expand [latex]{\left(x+y\right)}^{n}[/latex] by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand [latex]{\left(x+y\right)}^{52}[/latex], we might multiply [latex]\left(x+y\right)[/latex] by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

[latex]\begin{array}{l}{\left(x+y\right)}^{2}={x}^{2}+2xy+{y}^{2}\hfill \\ {\left(x+y\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}\hfill \\ {\left(x+y\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\hfill \end{array}[/latex]

First, let’s examine the exponents. With each successive term, the exponent for [latex]x[/latex] decreases and the exponent for [latex]y[/latex] increases. The sum of the two exponents is [latex]n[/latex] for each term. Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:

[latex]\left(\begin{array}{c}n\\ 0\end{array}\right),\left(\begin{array}{c}n\\ 1\end{array}\right),\left(\begin{array}{c}n\\ 2\end{array}\right),...,\left(\begin{array}{c}n\\ n\end{array}\right)[/latex].

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

[latex]\begin{array}{ll}{\left(x+y\right)}^{n}\hfill & =\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{n-k}{y}^{k}\hfill \\ \hfill & ={x}^{n}+\left(\begin{array}{c}n\\ 1\end{array}\right){x}^{n - 1}y+\left(\begin{array}{c}n\\ 2\end{array}\right){x}^{n - 2}{y}^{2}+...+\left(\begin{array}{c}n\\ n - 1\end{array}\right)x{y}^{n - 1}+{y}^{n}\hfill \end{array}[/latex]

Another way to see the coefficients is to examine the expansion of a binomial in general form, [latex]x+y[/latex], to successive powers 1, 2, 3, and 4.

[latex]\begin{array}{l}{\left(x+y\right)}^{1}=x+y\hfill \\ {\left(x+y\right)}^{2}={x}^{2}+2xy+{y}^{2}\hfill \\ {\left(x+y\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}\hfill \\ {\left(x+y\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\hfill \end{array}[/latex]

Can you guess the next expansion for the binomial [latex]{\left(x+y\right)}^{5}?[/latex] Graph of the function f_2. Above is an illustration the following:
  • There are [latex]n+1[/latex] terms in the expansion of [latex]{\left(x+y\right)}^{n}[/latex].
  • The degree (or sum of the exponents) for each term is [latex]n[/latex].
  • The powers on [latex]x[/latex] begin with [latex]n[/latex] and decrease to 0.
  • The powers on [latex]y[/latex] begin with 0 and increase to [latex]n[/latex].
  • The coefficients are symmetric.
To determine the expansion on [latex]{\left(x+y\right)}^{5}[/latex], we see [latex]n=5[/latex], thus, there will be 5+1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of [latex]x[/latex], the pattern is as follows:
  • Introduce [latex]{x}^{5}[/latex], and then for each successive term reduce the exponent on [latex]x[/latex] by 1 until [latex]{x}^{0}=1[/latex] is reached.
  • Introduce [latex]{y}^{0}=1[/latex], and then increase the exponent on [latex]y[/latex] by 1 until [latex]{y}^{5}[/latex] is reached. [latex]{x}^{5},{x}^{4}y,{x}^{3}{y}^{2},{x}^{2}{y}^{3},x{y}^{4},{y}^{5}[/latex]
The next expansion would be

[latex]{\left(x+y\right)}^{5}={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}[/latex].

But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal's Triangle. Pascal's Triangle To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3rd row, flank the ends of the rows with 1’s, and add [latex]1+1[/latex] to find the middle number, 2. In the [latex]n\text{th}[/latex] row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it. To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form. Pascal's Triangle expanded to show the values of the triangle as x and y terms with exponents

A General Note: The Binomial Theorem

The Binomial Theorem is a formula that can be used to expand any binomial.

[latex]\begin{array}{ll}{\left(x+y\right)}^{n}\hfill & =\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{n-k}{y}^{k}\hfill \\ \hfill & ={x}^{n}+\left(\begin{array}{c}n\\ 1\end{array}\right){x}^{n - 1}y+\left(\begin{array}{c}n\\ 2\end{array}\right){x}^{n - 2}{y}^{2}+...+\left(\begin{array}{c}n\\ n - 1\end{array}\right)x{y}^{n - 1}+{y}^{n}\hfill \end{array}[/latex]

How To: Given a binomial, write it in expanded form.

  1. Determine the value of [latex]n[/latex] according to the exponent.
  2. Evaluate the [latex]k=0[/latex] through [latex]k=n[/latex] using the Binomial Theorem formula.
  3. Simplify.

Example: Expanding a Binomial

Write in expanded form.
  1. [latex]{\left(x+y\right)}^{5}[/latex]
  2. [latex]{\left(3x-y\right)}^{4}[/latex]

Answer:

  1. Substitute [latex]n=5[/latex] into the formula. Evaluate the [latex]k=0[/latex] through [latex]k=5[/latex] terms. Simplify.
    [latex-display]\begin{array}{ll}{\left(x+y\right)}^{5}\hfill & =\left(\begin{array}{c}5\\ 0\end{array}\right){x}^{5}{y}^{0}+\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}{y}^{1}+\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}+\left(\begin{array}{c}5\\ 3\end{array}\right){x}^{2}{y}^{3}+\left(\begin{array}{c}5\\ 4\end{array}\right){x}^{1}{y}^{4}+\left(\begin{array}{c}5\\ 5\end{array}\right){x}^{0}{y}^{5}\hfill \\ {\left(x+y\right)}^{5}\hfill & ={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}\hfill \end{array}[/latex-display]
  2. Substitute [latex]n=4[/latex] into the formula. Evaluate the [latex]k=0[/latex] through [latex]k=4[/latex] terms. Notice that [latex]3x[/latex] is in the place that was occupied by [latex]x[/latex] and that [latex]-y[/latex] is in the place that was occupied by [latex]y[/latex]. So we substitute them. Simplify.
    [latex]\begin{array}{ll}{\left(3x-y\right)}^{4}\hfill & =\left(\begin{array}{c}4\\ 0\end{array}\right){\left(3x\right)}^{4}{\left(-y\right)}^{0}+\left(\begin{array}{c}4\\ 1\end{array}\right){\left(3x\right)}^{3}{\left(-y\right)}^{1}+\left(\begin{array}{c}4\\ 2\end{array}\right){\left(3x\right)}^{2}{\left(-y\right)}^{2}+\left(\begin{array}{c}4\\ 3\end{array}\right){\left(3x\right)}^{1}{\left(-y\right)}^{3}+\left(\begin{array}{c}4\\ 4\end{array}\right){\left(3x\right)}^{0}{\left(-y\right)}^{4}\hfill \\ {\left(3x-y\right)}^{4}\hfill & =81{x}^{4}-108{x}^{3}y+54{x}^{2}{y}^{2}-12x{y}^{3}+{y}^{4}\hfill \end{array}[/latex]

Analysis of the Solution

Notice the alternating signs in part b. This happens because [latex]\left(-y\right)[/latex] raised to odd powers is negative, but [latex]\left(-y\right)[/latex] raised to even powers is positive. This will occur whenever the binomial contains a subtraction sign.

Try It

Write in expanded form.
  1. [latex]{\left(x-y\right)}^{5}[/latex]
  2. [latex]{\left(2x+5y\right)}^{3}[/latex]

Answer:

  1. [latex]{x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5}[/latex]
  2. [latex]8{x}^{3}+60{x}^{2}y+150x{y}^{2}+125{y}^{3}[/latex]

Using the Binomial Theorem to Find a Single Term

Expanding a binomial with a high exponent such as [latex]{\left(x+2y\right)}^{16}[/latex] can be a lengthy process. Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term. Note the pattern of coefficients in the expansion of [latex]{\left(x+y\right)}^{5}[/latex].

[latex]{\left(x+y\right)}^{5}={x}^{5}+\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y+\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}+\left(\begin{array}{c}5\\ 3\end{array}\right){x}^{2}{y}^{3}+\left(\begin{array}{c}5\\ 4\end{array}\right)x{y}^{4}+{y}^{5}[/latex]

The second term is [latex]\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y[/latex]. The third term is [latex]\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}[/latex]. We can generalize this result.

[latex]\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}[/latex]

A General Note: The (r+1)th Term of a Binomial Expansion

The [latex]\left(r+1\right)\text{th}[/latex] term of the binomial expansion of [latex]{\left(x+y\right)}^{n}[/latex] is:

[latex]\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}[/latex]

How To: Given a binomial, write a specific term without fully expanding.

  1. Determine the value of [latex]n[/latex] according to the exponent.
  2. Determine [latex]\left(r+1\right)[/latex].
  3. Determine [latex]r[/latex].
  4. Replace [latex]r[/latex] in the formula for the [latex]\left(r+1\right)\text{th}[/latex] term of the binomial expansion.

Example: Writing a Given Term of a Binomial Expansion

Find the tenth term of [latex]{\left(x+2y\right)}^{16}[/latex] without fully expanding the binomial.

Answer: Because we are looking for the tenth term, [latex]r+1=10[/latex], we will use [latex]r=9[/latex] in our calculations.

[latex]\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}[/latex]

[latex]\left(\begin{array}{c}16\\ 9\end{array}\right){x}^{16 - 9}{\left(2y\right)}^{9}=5\text{,}857\text{,}280{x}^{7}{y}^{9}[/latex]

Try It

Find the sixth term of [latex]{\left(3x-y\right)}^{9}[/latex] without fully expanding the binomial.

Answer: [latex]-10,206{x}^{4}{y}^{5}[/latex]

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