**Introduction to Binomial Expansion****Expanding a Binomial****Finding a Specific Term with Binomial Expansion****More Practice**

# Introduction to Binomial Expansion

You’ll probably have to learn how to expand polynomials to various degrees (powers) using what we call the **Binomial Theorem** or **Binomial Expansion **(or **Binomial Series**).

We use this when we want to expand (multiply out) the power of a binomial like \({{\left( {x+y} \right)}^{n}}\) into a sum with terms \(a{{x}^{b}}{{y}^{c}}\), where *b* and *c* are non-negative integers (and it turns out that *b* + *c* = *n*). A **perfect square trinomial** is a simple example: \({{\left( {x+y} \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}\). (The coefficients in this case are **1**, **2**, and **1**, respectively.)

It just turns out that the coefficient *a* in this expansion is equal to \(\left( {\begin{array}{*{20}{c}} n \\ c \end{array}} \right)\) (also written as \(\displaystyle {}_{n}{{C}_{c}}\)), where \(\left( {\begin{array}{*{20}{c}} n \\ c \end{array}} \right)\,\,=\,\,\frac{{n!}}{{c!\left( {n-c} \right)!}}\) (this is called the **binomial coefficient**). Remember that \(n!=n\left( {n-1} \right)\left( {n-2} \right)\,\,\,….\) (until you get to 1). (You can also get \(\displaystyle {}_{n}{{C}_{c}}\) on your graphing calculator. Type in what you want for *n*, then MATH PROB, and hit **3** or scroll to nCr, and then type *c *and then ENTER). \(\displaystyle {}_{n}{{C}_{c}}\) is actually the number of ways to choose *c* items out of *n* terms, where order doesn’t matter – also called the **Combination **function.

Here is the **Binomial Theorem **(also called** Binomial Formula **or** Binomial Identity**):

\({{\left( {x+y} \right)}^{n}}\,\,=\,\,\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right){{x}^{n}}{{y}^{0}}\,+\,\left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right){{x}^{{n-1}}}{{y}^{1}}\,+\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){{x}^{{n-2}}}{{y}^{2}}\,+……+\left( {\begin{array}{*{20}{c}} n \\ {n-1} \end{array}} \right){{x}^{1}}{{y}^{{n-1}}}\,+\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right){{x}^{0}}{{y}^{n}}\)

or

\({{\left( {x+y} \right)}^{n}}\,\,=\sum\limits_{{k=0}}^{n}{{\,\left( {\begin{array}{*{20}{c}} n \\ k \end{array}} \right){{x}^{{n-k}}}{{y}^{k}}}}\), which is the same as \(\sum\limits_{{k=0}}^{n}{{\,\left( {\begin{array}{*{20}{c}} n \\ k \end{array}} \right){{x}^{k}}{{y}^{{n-k}}}}}\)

See how the exponents of the ** x**’s are going down (from

*n*to

**0**), while the exponents of the

*’s are going up (from 0 to*

**y***n*)? And remember that

**anything raised to the 0**is just

**1**. And for a binomial raise to the “

*n*”, we have “

*n*+ 1” terms.

The coefficients can also be found using a **Pascal Triangle**, which starts with **1**, and is a triangle with all **1**’s on the outside. Then on the inside, add the two numbers above to get the next number down:

\(\begin{array}{c}1\\1\,\,\,\,\,1\\1\,\,\,\,\,\,2\,\,\,\,\,\,1\\1\,\,\,\,\,\,3\,\,\,\,\,\,3\,\,\,\,\,\,1\\1\,\,\,\,\,\,4\,\,\,\,\,\,6\,\,\,\,\,\,4\,\,\,\,\,\,1\\1\,\,\,\,\,\,5\,\,\,\,\,\,10\,\,\,\,\,10\,\,\,\,\,5\,\,\,\,\,\,1\\1\,\,\,\,\,\,6\,\,\,\,\,\,15\,\,\,\,\,20\,\,\,\,\,15\,\,\,\,\,6\,\,\,\,\,\,1\\……\text{and so on}…..\end{array}\) Which is actually (first four rows): \(\begin{array}{c}\left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)\\\left( {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right)\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)\\\left( {\begin{array}{*{20}{c}} 2 \\ 0 \end{array}} \right)\,\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 2 \\ 1 \end{array}} \right)\,\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right)\\\left( {\begin{array}{*{20}{c}} 3 \\ 0 \end{array}} \right)\,\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 3 \\ 1 \end{array}} \right)\,\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right)\,\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 3 \\ 3 \end{array}} \right)\\……\text{and so on}…..\end{array}\)

As an example of how to use the Pascal Triangle, start with the second row for \({{\left( {x+y} \right)}^{1}}=1x+1y\), so the coefficients are both **1**. When using the Pascal Triangle, the exponent of the binomial is off by **1**; for example, we used the **2**^{nd} row to get the coefficients for \({{\left( {x+y} \right)}^{1}}\).

Here’s another illustration of just how Pascal’s Triangle is used for expanding binomials:

\(\displaystyle \begin{array}{l}{{\left( {x+y} \right)}^{0}}=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\{{\left( {x+y} \right)}^{1}}=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+y\\{{\left( {x+y} \right)}^{2}}=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+2xy+{{y}^{2}}\\{{\left( {x+y} \right)}^{3}}=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}+{{y}^{3}}\,\,\,\,\,\\{{\left( {x+y} \right)}^{4}}=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{4}}+4{{x}^{3}}y+6{{x}^{2}}{{y}^{2}}+4x{{y}^{3}}+{{y}^{4}}\,\,\,\,\\{{\left( {x+y} \right)}^{5}}=\,\,\,\,\,\,\,\,\,\,{{x}^{5}}+5{{x}^{4}}y+10{{x}^{3}}{{y}^{2}}+10{{x}^{2}}{{y}^{3}}+5x{{y}^{4}}+{{y}^{5}}\,\\{{\left( {x+y} \right)}^{6}}={{x}^{6}}+6{{x}^{5}}y+15{{x}^{4}}{{y}^{2}}+20{{x}^{3}}{{y}^{3}}+15{{x}^{2}}{{y}^{4}}+6x{{y}^{5}}+{{y}^{6}}\,\\\,\,…..\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}\)

Here’s a hint: when finding the coefficients of a binomial expansion using Pascal’s triangle, find the line with the **second term** the same as the power you want. For example, for a binomial with power **5**, use the line 1 **5** 10 10 5 1 for coefficients.

# Expanding a Binomial

The best way to show how Binomial Expansion works is to use an example. Let’s expand \({{\left( {x+3} \right)}^{6}}\) using the formula above. Here, the “*x*” in the generic binomial expansion equation is “*x*” and the “*y*” is “**3**”:

\(\begin{align}{{\left( {x+y} \right)}^{n}}\,\,&=\,\,\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right){{x}^{n}}{{y}^{0}}\,+\,\left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right){{x}^{{n-1}}}{{y}^{1}}\,+\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){{x}^{{n-2}}}{{y}^{2}}\,+……+\left( {\begin{array}{*{20}{c}} n \\ {n-1} \end{array}} \right){{x}^{1}}{{y}^{{n-1}}}\,+\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right){{x}^{0}}{{y}^{n}}\\\color{#804040}{{{{{\left( {x+3} \right)}}^{6}}\,}}\,&=\,\,\left( {\begin{array}{*{20}{c}} 6 \\ 0 \end{array}} \right){{x}^{6}}{{\left( 3 \right)}^{0}}\,+\,\left( {\begin{array}{*{20}{c}} 6 \\ 1 \end{array}} \right){{x}^{5}}{{\left( 3 \right)}^{1}}\,+\left( {\begin{array}{*{20}{c}} 6 \\ 2 \end{array}} \right){{x}^{4}}{{\left( 3 \right)}^{2}}\,+\left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right){{x}^{3}}{{\left( 3 \right)}^{3}}\,+\,\left( {\begin{array}{*{20}{c}} 6 \\ 4 \end{array}} \right){{x}^{2}}{{\left( 3 \right)}^{4}}+\,\left( {\begin{array}{*{20}{c}} 6 \\ 5 \end{array}} \right){{x}^{1}}{{\left( 3 \right)}^{5}}+\,\left( {\begin{array}{*{20}{c}} 6 \\ 6 \end{array}} \right){{x}^{0}}{{\left( 3 \right)}^{6}}\\&=\,\,1{{x}^{6}}{{\left( 3 \right)}^{0}}\,+\,6{{x}^{5}}{{\left( 3 \right)}^{1}}\,+15{{x}^{4}}{{\left( 3 \right)}^{2}}\,+20{{x}^{3}}{{\left( 3 \right)}^{3}}\,+\,15{{x}^{2}}{{\left( 3 \right)}^{4}}+\,6{{x}^{1}}{{\left( 3 \right)}^{5}}+\,1{{x}^{0}}{{\left( 3 \right)}^{6}}\\&=\,\,{{x}^{6}}\,+\,6{{x}^{5}}\left( 3 \right)\,+15{{x}^{4}}\left( 9 \right)\,+20{{x}^{3}}\left( {27} \right)\,+\,15{{x}^{2}}\left( {81} \right)+\,6{{x}^{1}}\left( {243} \right)+\,729\\&=\,\,{{x}^{6}}\,+\,18{{x}^{5}}\,+135{{x}^{4}}\,+540{{x}^{3}}\,+\,1215{{x}^{2}}+\,1458x+\,729\end{align}\)

Notice how the power (exponent) of the first variable starts at the highest (*n*) and goes down to **0** (which means that variable “disappears”, since \({{\left( {\text{anything}} \right)}^{0}}=1\)). Also notice that the power of the second variable starts at 0 (which means you don’t see it), and goes up to *n*.

Also notice that for the coefficients of the \(\left( {\begin{array}{*{20}{c}} {} \\ {} \end{array}} \right)\) part, the 6 (since this is *n*) always stays on top, and the bottom starts with 0 and goes up to 6. The exponents for the first term of the binomial with **6** (*n*) and goes down to **0**, and the exponent on the second term is always the bottom part of the \(\left( {\begin{array}{*{20}{c}} {} \\ {} \end{array}} \right)\). And if you add the two exponents, you always get **6** (since this is *n*).

Again, for the binomial coefficient \(\displaystyle \left( {\begin{array}{*{20}{c}} n \\ c \end{array}} \right)\), you can just use the \(\displaystyle {}_{n}{{C}_{c}}\) on your graphing calculator. (Type in what you want for *n*, then MATH PROB, and hit 3 or scroll to nCr, and then type *c* and then ENTER). You can also do these “by hand” by using \(\left( {\begin{array}{*{20}{c}} n \\ c \end{array}} \right)\,\,=\,\,\frac{{n!}}{{c!\left( {n-c} \right)!}}\). Notice that \(\displaystyle \left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right)\,\,\,\text{and}\,\,\,\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right)\) is always just 1 (0! = 1), and \(\displaystyle \left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right)\,\,\,\text{and}\,\,\,\left( {\begin{array}{*{20}{c}} n \\ {n-1} \end{array}} \right)\) is just *n*.

To use the **Pascal Triangle** above to do this, let’s look at the **7**^{th} row (since the first row is just “**1**”) to get the coefficients: 1 6 15 20 15 6 1. Note that since we’re wanting the **6**^{th} power, we are using the line that has **6** as the second term!

Let’s try another expansion, that’s a little more complicated. Here, the “*x*” in the generic binomial expansion equation is “4*a*” and the “*y*” is “–3*b*”:

\(\begin{align}{{\left( {x+y} \right)}^{n}}\,\,&=\,\,\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right){{x}^{n}}{{y}^{0}}\,+\,\left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right){{x}^{{n-1}}}{{y}^{1}}\,+\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){{x}^{{n-2}}}{{y}^{2}}\,+……+\left( {\begin{array}{*{20}{c}} n \\ {n-1} \end{array}} \right){{x}^{1}}{{y}^{{n-1}}}\,+\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right){{x}^{0}}{{y}^{n}}\\\color{#804040}{{{{{\left( {4a-3b} \right)}}^{4}}\,}}\,&=\,\,\left( {\begin{array}{*{20}{c}} 4 \\ 0 \end{array}} \right){{\left( {4a} \right)}^{4}}{{\left( {-3b} \right)}^{0}}\,+\,\left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right){{\left( {4a} \right)}^{3}}{{\left( {-3b} \right)}^{1}}\,+\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right){{\left( {4a} \right)}^{2}}{{\left( {-3b} \right)}^{2}}\,+\left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right){{\left( {4a} \right)}^{1}}{{\left( {-3b} \right)}^{3}}\,+\,\left( {\begin{array}{*{20}{c}} 4 \\ 4 \end{array}} \right){{\left( {4a} \right)}^{0}}{{\left( {-3b} \right)}^{4}}\\&=\,\,1\left( {256} \right){{a}^{4}}\,+\,4\left( {64} \right){{a}^{3}}\left( {-3b} \right)\,+6\left( {16{{a}^{2}}} \right)\left( {9{{b}^{2}}} \right)\,+4\left( {4a} \right)\left( {-27{{b}^{3}}} \right)\,+\,1\left( {81} \right){{b}^{4}}\\&=\,\,256{{a}^{4}}\,-768{{a}^{3}}b\,+864{{a}^{2}}{{b}^{2}}\,-432a{{b}^{3}}\,+\,81{{b}^{4}}\end{align}\)

We also could have used the **5**^{th} row of the Pascal Triangle to get the coefficients.

Notice how every other term is negative, since the second term of the binomial is negative.

# Finding Specific Terms with Binomial Expansion

You may be asked to find **specific terms** using the Binomial Expansion; for example, they may ask to find the **5**^{th} term of a binomial raised to an exponent, or the term containing say a certain variable raised to a power.

To do these, just remember that the *x*^{th} term has (*x* – 1) in the bottom of the \(\left( {\begin{array}{*{20}{c}} {} \\ {} \end{array}} \right)\) part of the **binomial coefficient**, since the first term has the \(\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right)\) part. (So the *x*^{th} term’s coefficient of a binomial expanded to the *n*th term is \(\left( {\begin{array}{*{20}{c}} n \\ {x-1} \end{array}} \right)\).)

Then remember that the exponent of the first part of the expanded terms is the difference of the two numbers in the \(\left( {\begin{array}{*{20}{c}} {} \\ {} \end{array}} \right)\), and the exponent of the second part of the expanded terms equals the bottom number in the \(\left( {\begin{array}{*{20}{c}} {} \\ {} \end{array}} \right)\) (since the two exponents always add up to equal *n*).

For example, if we are expanding a binomial raised to the **5**^{th} power, the **4**^{th} term will have a \(\left( {\begin{array}{*{20}{c}} 5 \\ {4-1} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right)\) coefficient, the power of the first expanded term **5 – 3 = 2**, and the power of the second is **3**.

Here are some examples. And remember that sometimes you will see \({}_{n}C{}_{r}\) instead of \(\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right)\):

Problem |
Explanation |

Find the 4^{th} term of the expansion of
\({{\left( {3x+5y} \right)}^{8}}\) |
The 4^{th} term will have a \(\left( {\begin{array}{*{20}{c}} 8 \\ 3 \end{array}} \right)\) binomial coefficient, since n = 8, and 4 – 1 = 3. The first exponent will be 8 – 3 = 5, and the second exponent will be 3. The 4^{th} term then is:
\(\left( {\begin{array}{*{20}{c}} 8 \\ 3 \end{array}} \right){{\left( {3x} \right)}^{5}}{{\left( {5y} \right)}^{3}}=56\cdot 243{{x}^{5}}\cdot 125{{y}^{3}}=1701000{{x}^{5}}{{y}^{3}}\) |

Find the 5^{th} term of the expansion of
\({{\left( {\sqrt{a}-\sqrt{b}} \right)}^{7}}\) |
The 5^{th} term will have a \(\left( {\begin{array}{*{20}{c}} 7 \\ 4 \end{array}} \right)\) binomial coefficient, since n = 7, and 5 – 1 = 4. The first exponent will be 7 – 4 = 3, and the second exponent will be 4. The 5^{th} term then is:
\(\left( {\begin{array}{*{20}{c}} 7 \\ 4 \end{array}} \right){{\left( {\sqrt{a}} \right)}^{3}}{{\left( {-\sqrt{b}} \right)}^{4}}=35\cdot {{\left( {{{a}^{{\frac{1}{2}}}}} \right)}^{3}}\cdot {{\left( {-{{b}^{{\frac{1}{2}}}}} \right)}^{4}}=35\cdot {{a}^{{\frac{3}{2}}}}\cdot {{b}^{2}}=35a\sqrt{a}{{b}^{2}}\) |

Find the term containing \({{x}^{6}}\) in the expansion of
\({{\left( {2x-2y} \right)}^{{10}}}\) |
The term containing \({{x}^{6}}\) will have a 6^{th} power in the first part of the term, or \({{\left( {2x} \right)}^{6}}\). Since n = 10, this means the binomial coefficient will have to be 4, since the exponent on the first part is the difference of the numbers in the binomial coefficient: 10 – 4 = 6 (this will be the 5^{th} term). Now we can “build” the rest of the term:
\(\displaystyle \left( {\begin{array}{*{20}{c}} {10} \\ 4 \end{array}} \right){{\left( {2x} \right)}^{6}}{{\left( {-2y} \right)}^{4}}=210\cdot 64{{x}^{6}}\cdot 16{{y}^{4}}=215040{{x}^{6}}{{y}^{4}}\) Notice that the negative goes away when we raise to an even exponent. |

Find the term in the expansion of
\({{\left( {x-y} \right)}^{6}}\) where |
The term where x and y are the same must have an \({{x}^{3}}{{y}^{3}}\) in it, since the two exponents must add up to 6 (n). Then the binomial coefficient must be \(\left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right)\), since n = 6, and 6 – 3 must equal the first power (3). Now we can “build” the rest of the term:
\(\displaystyle \left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right){{x}^{3}}{{\left( {-y} \right)}^{3}}=20\cdot {{x}^{3}}\cdot -{{y}^{3}}=-20{{x}^{3}}{{y}^{3}}\) |

Find the term containing \({{b}^{4}}\) in the expansion of
\({{\left( {2a-5b} \right)}^{7}}\) |
This one’s a little tricky, since b in the second part of the binomial. When b is raised to the 4^{th} power, we must have a 4 on the bottom of the binomial coefficient and we always have n (7) on the top. (This will be the 5^{th} term). So this isn’t so bad! Now we can “build” the rest of the term:
.\(\displaystyle \left( {\begin{array}{*{20}{c}} 7 \\ 4 \end{array}} \right){{\left( {2a} \right)}^{3}}{{\left( {-5b} \right)}^{4}}=35\cdot 8{{a}^{3}}\cdot 625{{b}^{4}}=175000{{a}^{3}}{{b}^{4}}\) |

Find the term containing \({{y}^{8}}\) in the expansion of
\({{\left( {3{{y}^{2}}-2{{z}^{4}}} \right)}^{5}}\) |
Since we want the 8^{th} power of y, we must raise \(3{{y}^{2}}\) to the 4^{th} power, since when we raise an exponent to another exponent, we multiply exponents. This means, for the binomial coefficient, we’ll have \(\left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right)\), since n = 5, and 5 – 1 = 4 (this will be the 2^{nd} term). Now we can “build” the rest of the term:
. \(\displaystyle \left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right){{\left( {3{{y}^{2}}} \right)}^{4}}{{\left( {-2{{z}^{4}}} \right)}^{1}}=5\cdot 81{{y}^{8}}\cdot -2{{z}^{4}}=-810{{y}^{8}}{{z}^{4}}\) It works! |

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