7.12: Binomial Theorem and Expansions
Multiplying binomials is not a terribly difficult exercise, but it can certainly be time consuming with higher exponents, for example:

Calculate: \begin{align*}(x  3)^2\end{align*} is pretty easy:
 \begin{align*}(x  3) \cdot (x  3)\end{align*}
 \begin{align*}(x^2 6x +9)\end{align*}
...not bad
 BUT...

Calculate: \begin{align*}(x  3)^5\end{align*}
 This is: \begin{align*}(x  3) \cdot (x  3) \cdot (x  3) \cdot (x  3) \cdot (x  3)\end{align*}
 First we "foil" the first two terms to get \begin{align*}(x^2 6x +9)\end{align*}
 2nd, we multiply \begin{align*}(x^2 6x +9)\end{align*} by \begin{align*}(x  3)\end{align*}, yielding: \begin{align*}(x^3 9x^2 +27x 27)\end{align*}
 3rd, we multiply \begin{align*}(x^3 9x^2 +27x 27)\end{align*} by \begin{align*}(x  3)\end{align*}
... and so on.
Definitely doable, but a nightmare of a job, particularly by hand.
Isn't there an easier way?
Watch This
Embedded Video:
 James Sousa: Binomial Expansion using Pascal's Triangle
Guidance
There is a particular pattern in combinations that is seen in the expansion of polynomials of the form (x + y)^{n}.
This pattern is most commonly displayed in a triangle:
This triangle is referred to as Pascal’s triangle, named after mathematician Blaise Pascal, although other mathematicians before him worked with these numbers. The numbers in the triangle can be used to generate more rows: notice that if you add two consecutive numbers, you get the number between and below them in the next row.
We can generalize this pattern as follows: \begin{align*}\binom{n} {r  1} + \binom{n} {r} = \binom{n + 1} {r}\end{align*}.
Binomial Expansion
To expand a binomial is to multiply all of the factors. The resulting polynomial is in standard form. For example:
(x + y)^{2} = (x + y) (x + y) = x^{2} + xy + xy + y^{2} = x^{2} + 2xy + y^{2}
If we expand (x + y)^{3} , we get:
(x + y)^{3}  = (x + y) (x + y) (x + y)  

= (x + y) (x^{2} + 2xy + y^{2})  
= x^{3} + 2x^{2}y + xy^{2} + x^{2}y + 2xy^{2} + y^{3}  
= x^{3} + 3x^{2}y + 3xy^{2} + y^{3} 

 Notice that the coefficients of each polynomial correspond to a row of Pascal’s triangle.
Also notice that the exponents of x descend, and the exponents of y ascend with each term. These are key aspects of the Binomial Theorem.
The Binomial Theorem
The binomial theorem can be stated using a summation:
\begin{align*}(x + y)^n = \sum_{r = 0}^n \left (\binom{n} {r} x^{n  r} y^r \right)\end{align*} 

This is a very succinct way of summarizing the pattern in a binomial expansion. Let’s return to (x + y)^{3} to see how the theorem works.
\begin{align*}(x + y)^3\end{align*}  \begin{align*}= \binom{3} {0}x^{3  0} y^0 + \binom{3} {1}x^{3 1}y^1 + \binom{3} {2}x^{3  2}y^2 + \binom{3} {3}x^{3 3}y^3\end{align*}  

\begin{align*}= 1x^3 \cdot 1 + 3x^2y + 3xy^2 + 1 \cdot x^0 \cdot y^3\end{align*}  
\begin{align*}= x^3 + 3x^2y + 3xy^2 + y^3\end{align*} 
Again, the exponents on x descend from 3 to 0. The exponents on y ascend from 0 to 3. The coefficients on the terms correspond to row 3 of Pascal’s triangle. These coefficients are, not surprisingly, referred to as binomial coefficients!
Given this theorem, we can expand any binomial without having to multiply all of the factors.
Finding a specific term in a Binomial Expansion
Finding a term in an expansion can be used to answer a particular kind of probability question.
Consider an experiment, in which there are two possible outcomes, such as flipping a coin. If we flip a coin over and over again, this is referred to as a Bernoulli trial. In each flip (“experiment”), the probability of getting a head is 0.5, and the probability of getting a tail is 0.5. (Note: this is true for flipping a coin, but not for other situations. That is, it’s not always “5050 chance!) Now say we flip a coin 25 times. What is the probability of getting exactly 10 heads?
The answer to this question is a term of a binomial expansion. That is, the probability of getting 10 heads from 25 coin tosses is: \begin{align*} \binom{25}{10} (0.5)^{10}(0.5)^{15}\approx 0.0974\end{align*}, or about a 10% chance.
Example A
Use the binomial theorem to expand each polynomial:

a.(2x + a)^{4} b. (x  3)^{5}
Solution
 a.(2x + a)^{4}
= 1(2x)^{4} (a)^{0} + 4(2x)^{3} (a)^{1} + 6(2x)^{2}(a)^{2} + 1(2x)^{0} (a)^{4}  

= 16x^{4} + 32x^{3}a + 24x^{2}a^{2} + 8x'a^{3} + a^{4} 

 Note that it is easier to simply use the numbers from the appropriate row of the triangle than to write out all of the coefficients as combinations. However, if n is large, it may be easier to use the combinations.
 b. (x  3)^{5}
= 1(x)^{5} (3)^{0} + 5(x)^{4} (3)^{1} + 10(x)^{3}(3)^{2} + 10(x)^{2}(3)^{3} + 5(x)^{1} (3)^{4} + 1(x)^{0} (3)^{5}  

= x^{5}  1x^{4}a + 90x^{3}  270x^{2} + 405x  243 

 Notice that in this expansion, the terms alternate signs. This is the case because the second term in the binomial is 3. When expanding this kind of polynomial, be careful with your negatives!
We can also use the Binomial Theorem to identify a particular term or coefficient.
Example B
Identify the 3^{rd} term of the expansion of (2x + 3)^{6}.
Solution
The 3^{rd} term is \begin{align*}\mathit \ \binom{6}{2} (2x)^{4}3^{2}=15\cdot 16x^{4} \cdot 9 = 2160x^{4}\end{align*}
 Keep in mind that row 6 of Pascal’s triangle starts with \begin{align*}\mathit \ \binom{6}{0}\end{align*} , so the coefficient of the third term in the expansion is \begin{align*}\mathit \ \binom{6}{2}\end{align*}. Also keep in mind that the first term includes (2x)^{6} and 3^{0} , so the third term includes (2x)^{4} and 3^{2}.
Example C
What is the coefficient of \begin{align*}x^2\end{align*} in the expansion of \begin{align*}(3x +3)^3\end{align*}
Solution
Use the binomial theorem for the second term:
 \begin{align*}_3C_2(3x)^2(3)^{32}\end{align*}
 \begin{align*}3(3x)^2(3)^1 = 81\end{align*}
Vocabulary
The Binomial Theorem is an efficient formula for calculating the expansion of binomials.
Binomial Expansion is the process of raising a binomial such as (x + 2) to a power. The process can be time consuming, particularly with higher exponents.
Pascal's Triangle is a pyramid of sorts constructed with the coefficients of binomials as they are expanded. It is a convenient reference.
Guided Practice
1) Use Pascal's triangle to find the coefficient of \begin{align*}x^2y\end{align*} in \begin{align*}(x  y)^3\end{align*}
2) Expand: \begin{align*}(x + 3)^3\end{align*}
3) Expand: \begin{align*}(3x  3)^6\end{align*}
4) Find the coefficient of\begin{align*}x^3\end{align*} in \begin{align*}(x  y)^3\end{align*}
Solutions
1) First, create Pascal's triangle. Remember to add 1's on the ends and the sum of the two numbers above to get the new numbers.


 \begin{align*}1\end{align*}


 \begin{align*}1  1\end{align*}
 \begin{align*}1  2  1\end{align*}
\begin{align*}1  3  3  1\end{align*}
We only had to go up to the third row, we are looking for the 2nd term, so we count over \begin{align*} 1+ 1\end{align*} from the left to get our coefficient of 3. Don't forget to start with the 1st term.
We use the formula for the second term:
\begin{align*}_3C_2 x^2(y)^{32}\end{align*}
\begin{align*}3x^2(y)^1\end{align*}
So we get an answer of: 3
2) Start by expanding out each term separately:
Use the formula for term 0:
 \begin{align*}_3C_3x^3(3)^{33} = x^3(3)^0 = x ^3\end{align*}
...and term 1:
 \begin{align*}_3C_2x^2(3)^{32} = 3x^2(3)^1 = 9x ^2\end{align*}
...and term 2:
 \begin{align*}_3C_1x^1(3)^{31} = 3x^1(3)^2 = 27x \end{align*}
...and term 3:
 \begin{align*}_3C_0x^0(3)^{30} = x^0(3)^3 = 27\end{align*}
So we end up with:
 \begin{align*}x^3 + 9x^2 + 27x + 27\end{align*}
3) We will expand out each term separately:
We use the formula for term 0:
 \begin{align*}_6C_6 (3x)^6(3)^{6  6} = (3x)^6(3)^0 = 729x^6\end{align*}
...and term 1:
 \begin{align*}_6C_5 (3x)^5(3)^{6  5} = 6(3x)^5(3)^1 = 4374x^5\end{align*}
...and term 2:
 \begin{align*}_6C_4 (3x)^4(3)^{6  4} = 15(3x)^4(3)^2 = 10,935x^4\end{align*}
...and term 3:
 \begin{align*}_6C_3 (3x)^3(3)^{6  3} = 20(3x)^3(3)^3 = 14,580x^3\end{align*}
...and term 4:
 \begin{align*}_6C_2 (3x)^2(3)^{6  2} = 15(3x)^2(3)^4 = 10,935x^2\end{align*}
...and term 5:
 \begin{align*}6C_1 (3x)^1(3)^{6  1} = 6(3x)^1(3)^5 = 4374x\end{align*}
...and term 6:
 \begin{align*}_6C_0 (3x)^0(3)^{6  0} = (3x)^0(3)^6 = 729\end{align*}
 The expanded polynomial is:
 \begin{align*}729x^6  4374x^5 + 10,935x^4  14,580x^3 + 10,935x^2  4374x + 729\end{align*}
4) We use the formula for the third term:
 \begin{align*}_3C_3 x^3(y)^{3  3}\end{align*}
 \begin{align*}x^3(y)^0\end{align*}
 Giving us our answer: 1
Practice
Questions
 Expand: \begin{align*}(x+3a)^{4}\end{align*}
 Expand: \begin{align*}(y + \frac{1}{2})^5\end{align*}
 Expand: \begin{align*}(2xa)^{6}\end{align*}
 Expand: \begin{align*}(x + y)^6\end{align*}
 Expand: \begin{align*}(3x + 1)^5\end{align*}
 Expand: \begin{align*}(x + y)^5\end{align*}
 Expand: \begin{align*} (2x + 2)^4\end{align*}
Find the Term
 Find the 3^{rd} term in the expansion \begin{align*}(3x+2a)^{9}\end{align*}.
 Find the 7^{th} term in the expansion of \begin{align*}(4x  \frac{1}{2}a)^{10}\end{align*}.
Use Pascal's Triangle to find the coefficient:
 What is the coefficient of \begin{align*}x^3y^2\end{align*} in the expansion of \begin{align*}(x + y)^5\end{align*}?
 What is the coefficient of \begin{align*}x^4\end{align*} in the expansion\begin{align*}(3x + 1)^4\end{align*}?
 What is the coefficient of \begin{align*}x\end{align*} in the expansion:\begin{align*}(2x + 2)^5\end{align*}?
What is the coefficient of the expansions?
 What is the coefficient of \begin{align*}x^2\end{align*} in the expansion of \begin{align*}(x + 1) ^6\end{align*}?
 What is the coefficient of \begin{align*}x^5\end{align*} in the expansion of \begin{align*}(2x + 1) ^5\end{align*}?
 What is the coefficient of \begin{align*}x\end{align*} in the expansion of \begin{align*}(3x + 2) ^3\end{align*}?
 What is the coefficient of \begin{align*}x^6\end{align*} in the expansion of \begin{align*}(x + y) ^6\end{align*}?
 What is the coefficient of \begin{align*}x\end{align*} in the expansion of \begin{align*}(2x + 1) ^6\end{align*}?
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In probability, a Bernoulli trial is an experiment with exactly two possible outcomesBinomial Expansion
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