Binomial Expansion

Introduction to Binomial Expansion

The Binomial Theorem or Binomial Expansion (or Binomial Series) is a formula used to expand binomials 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 $ 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 turns out that the coefficient $ a$ in the terms $ a{{x}^{b}}{{y}^{c}}$ of this expansion is equal to $ \left( {\begin{array}{*{20}{c}} n \\ c \end{array}} \right)$ (also written as $ \displaystyle {}_{n}{{C}_{c}}$), where $ \displaystyle \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$ factorial, or $ 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 (or go directly to nCr and type in both numbers). $ \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 that we also saw here in the Introduction to Statistics and Probability section.

Thus, here is the formula for the Binomial Theorem (also called the 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

Summation Notation: $ {{\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 $ y$’s are going up (from 0 to $ n$)? And remember that anything raised to the 0 is just 1. Note that for a binomial raise to the “$ n$”, there are “$ n+1$” terms.

Pascal Triangle

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. On the inside, the next number down is the sum of the two numbers above:

$ \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, the 3nd row shows the coefficients for $ {{\left( {x+y} \right)}^{2}}={{x}^{2}}+2x+{{y}^{2}}$, with coefficients 1  2  1. Note that when using the Pascal Triangle, the exponent of the binomial is off by 1; for example, we used the 3nd row to get the coefficients for $ {{\left( {x+y} \right)}^{2}}$.

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. 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$) always stays on top, and the bottom starts with 0 and goes up to 6. The exponents for the first term of the binomial start with 6 ($ 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)$. If you add the two exponents, you always get 6, which 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 $ \displaystyle \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 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 $ n$.

To use the Pascal Triangle above to do this, look at the 7th row (since the first row is just “1”) to get the coefficients:   1   6   15   20   15   6   1. Note that since we want the 6th power, use the line that has 6 as the second term!

Here are a few more that are a little more complicated, including one that’s a Complex Number:

Binomial Expansion Problem Explanation
Expand the binomial:

 

$ {{\left( {4a-3b} \right)}^{4}}$

 

 

Here, the “$ x$” in the generic binomial expansion equation is “$ 4a$” and the “$ y$” is “$ -3b$”.

$ \displaystyle \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{#800000}{{{{{\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 5th 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.

Expand the binomial:

 

$ {{\left( {1-4i} \right)}^{5}}$

 

$ \displaystyle \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{#800000}{{{{{\left( {1-4i} \right)}}^{5}}}}&=\left( {\begin{array}{*{20}{c}} 5 \\ 0 \end{array}} \right){{\left( 1 \right)}^{5}}{{\left( {-4i} \right)}^{0}}+\left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right){{\left( 1 \right)}^{4}}{{\left( {-4i} \right)}^{1}}+\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right){{\left( 1 \right)}^{3}}{{\left( {-4i} \right)}^{2}}\,+\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right){{\left( 1 \right)}^{2}}{{\left( {-4i} \right)}^{3}}\,\\&\,\,\,\,\,+\left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right){{\left( 1 \right)}^{1}}{{\left( {-4i} \right)}^{4}}+\left( {\begin{array}{*{20}{c}} 5 \\ 5 \end{array}} \right){{\left( 1 \right)}^{0}}{{\left( {-4i} \right)}^{5}}\\&=1\left( 1 \right)\left( 1 \right)\,+\,5\left( 1 \right)\left( {-4i} \right)\,+10\left( 1 \right)\left( {-16} \right)\,+10\left( 1 \right)\left( {64i} \right)\,+\,5\left( 1 \right)\left( {256} \right)+1\left( 1 \right)\left( {-1024i} \right)\\&=1\,-20i\,-160\,+640i\,+\,1280-1024i\\&=1121-404i\end{align}$

(We also could have used the 6th row of the Pascal Triangle to get the coefficients.)

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 5th term of a binomial raised to an exponent, or the term containing 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. 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)$. 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, to expand a binomial raised to the 5th power, the 4th term has 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. Also remember that sometimes you will see $ {}_{n}C{}_{r}$ instead of $ \left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right)$:

Binomial Expansion Problem Explanation
Find the 4th term of the expansion of 

$ {{\left( {3x+5y} \right)}^{8}}$

The 4th term has a $ \left( {\begin{array}{*{20}{c}} 8 \\ 3 \end{array}} \right)$ binomial coefficient, since $ n=8$ and $ 4-1=3$. The first exponent is $ 8-3=5$,  and the second exponent is 3. Thus, the 4th term 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 5th term of the expansion of

$ {{\left( {\sqrt{a}-\sqrt{b}} \right)}^{7}}$

The 5th term has a $ \left( {\begin{array}{*{20}{c}} 7 \\ 4 \end{array}} \right)$ binomial coefficient, since $ n=7$ and $ 5-1=4$. The first exponent is $ 7-4=3$, and the second exponent is 4. Thus, the 5th term 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}}$ has a 6th power in the first part of the term, or $ {{\left( {2x} \right)}^{6}}$. Since $ n=10$, the binomial coefficient has to be 4, since the exponent on the first part is the difference of the numbers in the binomial coefficient: $ 10-4=6$; this is the 5th term. Now, “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 $ x$ and $ y$ have the same power.

The term where $ x$ and $ y$ have the same power must have an $ {{x}^{3}}{{y}^{3}}$ in it, since the two exponents must add up to 6 ($ n$). 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). ; this is the 4th term. Now, “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$ is in the second part of the binomial. When $ b$ is raised to the 4th power, there must be a 4 on the bottom of the binomial coefficient and an $ n$ (7) on the top; this is the 5th term. Now, “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 8th power of $ y$, raise $ 3{{y}^{2}}$ to the 4th power, since when raising 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 is the 2nd term. Now, “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!

Learn these rules, and practice, practice, practice!


Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Introduction to Limits – you are ready!