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Binomial Theorem and Expansions

Expansion of binomials raised to a power using combinations.

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Binomial Theorem

The Binomial Theorem tells you how to expand a binomial such as \begin{align*}(2x-3)^5\end{align*} without having to compute the repeated distribution. What is the expanded version of \begin{align*}(2x-3)^5\end{align*}?

Introduction to the Binomial Theorem

The Binomial Theorem states:

\begin{align*}(a+b)^n=\sum \limits_{i=0}^n \tbinom{n}{i} a^i b^{n-i}\end{align*}

Writing out a few terms of the summation symbol helps you to understand how this theorem works:

\begin{align*}(a+b)^n=\tbinom{n}{0}a^n + \tbinom{n}{1}a^{n-1}b^1 + \tbinom{n}{2}a^{n-2}b^2 + \cdots + \tbinom{n}{n} b^n\end{align*}

Going from one term to the next in the expansion, you should notice that the exponents of \begin{align*}a\end{align*} decrease while the exponents of \begin{align*}b\end{align*} increase. You should also notice that the coefficients of each term are combinations. Recall that \begin{align*}\tbinom{n}{0}\end{align*} is the number of ways to choose  objects from a set of \begin{align*}n\end{align*} objects. 

Take the following binomial: 


It can be expanded using the Binomial Theorem:

\begin{align*}(m-n)^6 &=\tbinom{6}{0}m^6 + \tbinom{6}{1}m^5(-n)^1 + \tbinom{6}{2}m^4(-n)^2 + \tbinom{6}{3}m^3(-n)^3 \\ & \ \ \ + \tbinom{6}{4}m^2(-n)^4 + \tbinom{6}{5}m^1(-n)^5 + \tbinom{6}{6}(-n)^6 \\ &=1m^6-6m^5n+15m^4n^2-20m^3n^3+15m^2n^4-6m^1n^5+1n^6\end{align*}

Be extremely careful when working with binomials of the form \begin{align*}(a-b)^n\end{align*}. You need to remember to capture the negative with the second term as you write out the expansion: \begin{align*}(a-b)^n=(a+(-b))^n\end{align*}.

Another way to think about the coefficients in the Binomial Theorem is that they are the numbers from Pascal’s Triangle. Look at the expansions of \begin{align*}(a+b)^n\end{align*} below and notice how the coefficients of the terms are the numbers in Pascal’s Triangle.


Example 1

Earlier, you were asked to expand \begin{align*}(2x-3)^5\end{align*}The expanded version of \begin{align*}(2x-3)^5\end{align*}  is:

\begin{align*}(2x-3)^5 &=\tbinom{5}{0} (2x)^5+\tbinom{5}{1} (2x)^4(-3)^1+\tbinom{5}{2} (2x)^3(-3)^2\\ & \ \ \ +\tbinom{5}{3} (2x)^2(-3)^3+\tbinom{5}{4} (2x)^1(-3)^4+ \tbinom{5}{5}(-3)^6 \\ &=(2x)^5+5(2x)^4(-3)^1+10(2x)^3(-3)^2\\ & \ \ \ +10(2x)^2(-3)^3+5(2x)^1(-3)^4+(-3)^5 \\ &=32x^5-240x^4+720x^3-1080x^2+810x-243\end{align*}

Example 2

What is the coefficient of the term \begin{align*}x^7y^9\end{align*} in the expansion of the binomial \begin{align*}(x+y)^{16}\end{align*}?

The Binomial Theorem allows you to calculate just the coefficient you need.

\begin{align*}\tbinom{16}{9}=\frac{16!}{9!7!}=\frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=11,440\end{align*}

Example 3

What is the coefficient of \begin{align*}x^6\end{align*} in the expansion of \begin{align*}(4-3x)^7\end{align*}?

For this problem you should calculate the whole term, since the 3 and the 4 in \begin{align*}(3-4x)\end{align*} will impact the coefficient of \begin{align*}x^6\end{align*} as well. \begin{align*}\tbinom{7}{6}4^1(-3x)^6=7 \cdot 4 \cdot 729x^6=20,412x^6\end{align*}. The coefficient is 20,412.

Example 4

Compute the following summation.

\begin{align*}\sum \limits_{i=0}^4 \tbinom{4}{i}\end{align*}

This is asking for \begin{align*}\tbinom{4}{0}+\tbinom{4}{1}+\cdots+\tbinom{4}{4}\end{align*}, which are the sum of all the coefficients of \begin{align*}(a+b)^4\end{align*}.


Example 5

Collapse the following polynomial using the Binomial Theorem.


Since the last term is -1 and the power on the first term is a 5 you can conclude that the second half of the binomial is \begin{align*}(? -1)^5\end{align*}. The first term is positive and \begin{align*}(2x)^5=32x^5\end{align*}, so the first term in the binomial must be \begin{align*}2x\end{align*}. The binomial is \begin{align*}(2x-1)^5\end{align*}.


Expand each of the following binomials using the Binomial Theorem.

1. \begin{align*}(x-y)^4\end{align*}

2. \begin{align*}(x-3y)^5\end{align*}

3. \begin{align*}(2x+4y)^7\end{align*}

4. What is the coefficient of \begin{align*}x^4\end{align*} in \begin{align*}(x-2)^7\end{align*}?

5. What is the coefficient of \begin{align*}x^3 y^5\end{align*} in \begin{align*}(x+y)^8\end{align*}?

6. What is the coefficient of \begin{align*}x^5\end{align*} in \begin{align*}(2x-5)^6\end{align*}?

7. What is the coefficient of \begin{align*}y^2\end{align*} in \begin{align*}(4y-5)^4\end{align*}?

8. What is the coefficient of \begin{align*}x^2y^6\end{align*} in \begin{align*}(2x+y)^8\end{align*}?

9. What is the coefficient of \begin{align*}x^3y^4\end{align*} in \begin{align*}(5x+2y)^7\end{align*}?

Compute the following summations.

10. \begin{align*}\sum \limits_{i=0}^9 \tbinom{9}{i}\end{align*}

11. \begin{align*}\sum \limits_{i=0}^{12} \tbinom{12}{i}\end{align*}

12. \begin{align*}\sum \limits_{i=0}^{8} \tbinom{8}{i}\end{align*}

Collapse the following polynomials using the Binomial Theorem.

13. \begin{align*}243x^5-405x^4+270x^3-90x^2=15x-1\end{align*}

14. \begin{align*}x^7-7x^6y+21x^5y^2-35x^4y^3+35x^3y^4-21x^2y^5+7xy^6-y^7\end{align*}

15. \begin{align*}128x^7-448x^6y+672x^5y^2-560x^4y^3+280x^3y^4-84x^2y^5+14xy^6-y^7\end{align*}

Review (Answers)

To see the Review answers, open this PDF file and look for section 12.8. 

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Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.

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