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Binomial Theorem and Expansions

Practice Binomial Theorem and Expansions
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Binomial Theorem

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

Watch This

http://www.youtube.com/watch?v=YxysKtqpbVI James Sousa: The Binomial Theorem Using Combinations

http://www.youtube.com/watch?v=NLQmQGA4a3M James Sousa: The Binominal Theorem Using Pascal’s Triangle


The Binomial Theorem states:

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

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

(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

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

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 (a+b)^n  below and notice how the coefficients of the terms are the numbers in Pascal’s Triangle.

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

Example A

Expand the following binomial using the Binomial Theorem. 



(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

Example B

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

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

\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

Example C

What is the coefficient of  x^6  in the expansion of (4-3x)^7 ?

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

Concept Problem Revisited

The expanded version of (2x-3)^5   is:

(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


The Binomial Theorem is a theorem that states how to expand binomials that are raised to a power using combinations. The Binomial Theorem is:

(a+b)^n=\sum \limits_{i=0}^n \tbinom{n}{i}a^ib^{n-i}

Guided Practice

1. What is the coefficient of  x^3  in the expansion of  (x-4)^5 ?

2. Compute the following summation.

\sum \limits_{i=0}^4 \tbinom{4}{i}

3. Collapse the following polynomial using the Binomial Theorem.



1.  \tbinom{5}{2} \cdot 1^3(-4)^2=160

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


3. 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 (? -1)^5 . The first term is positive and (2x)^5=32x^5 , so the first term in the binomial must be 2x . The binomial is (2x-1)^5 .


Expand each of the following binomials using the Binomial Theorem.

1. (x-y)^4

2. (x-3y)^5

3. (2x+4y)^7

4. What is the coefficient of  x^4  in (x-2)^7 ?

5. What is the coefficient of x^3 y^5 in (x+y)^8 ?

6. What is the coefficient of  x^5  in (2x-5)^6 ?

7. What is the coefficient of y^2 in (4y-5)^4 ?

8. What is the coefficient of  x^2y^6  in (2x+y)^8 ?

9. What is the coefficient of  x^3y^4  in (5x+2y)^7 ?

Compute the following summations.

10. \sum \limits_{i=0}^9 \tbinom{9}{i}

11. \sum \limits_{i=0}^{12} \tbinom{12}{i}

12. \sum \limits_{i=0}^{8} \tbinom{8}{i}

Collapse the following polynomials using the Binomial Theorem.

13. 243x^5-405x^4+270x^3-90x^2=15x-1

14. x^7-7x^6y+21x^5y^2-35x^4y^3+35x^3y^4-21x^2y^5+7xy^6-y^7

15. 128x^7-448x^6y+672x^5y^2-560x^4y^3+280x^3y^4-84x^2y^5+14xy^6-y^7




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

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