This document discusses binomial expansion, which is a method for expanding binomials like (x + a)^n without lengthy multiplication. It introduces key concepts like Pascal's triangle for finding coefficients and the binomial theorem for determining the general pattern of terms in the expansion. Examples are worked through to demonstrate expanding specific binomials like (2x - 3y)^6 according to this method.

1. ( )
n
x a+

2. ( )
1
x a x a+ = +
A binomial is a polynomial with two terms such as x + a. Often we
need to raise a binomial to a power. In this section we'll explore a
way to do just that without lengthy multiplication.
( )
2 2 2
2x a x ax a+ = + +
( )
3 3 2 2 3
3 3x a x ax a x a+ = + + +
( )
4 4 3 2 2 3 4
4 6 4x a x ax a x a x a+ = + + + +
Can you see a
pattern?
Can you make a
guess what the next
one would be?
( )
5
x a+ = 5 4 2 3 3 2 4 5
__ __ __ __x ax a x a x a x a+ + + + +
We can easily see the pattern on the x's and the a's. But what
about the coefficients? Make a guess and then as we go we'll
see how you did.
( )
0
1x a+ =

3. ( )
1
1 1x a x a+ = +
Let's list all of the coefficients on the x's and the a's and look for a
pattern.
( )
2 2 2
1 2 1x a x ax a+ = + +
( )
3 3 2 2 3
1 3 3 1x a x ax a x a+ = + + +
( )
4 4 3 2 2 3 4
1 4 6 4 1x a x ax a x a x a+ = + + + +
( )
0
1x a+ =
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
+
+ +
+ + +
Can you
guess the
next row?
1 5 10 10 5 1
+ + + +
( )
5
x a+ = 5 4 2 3 3 2 4 5
1 5 10 10 5 1x ax a x a x a x a+ + + + +

4. 1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
This is called Pascal's Triangle and would give us the
coefficients for a binomial expansion of any power if we
extended it far enough.
This is good for
lower powers but
could get very
large. We will
introduce some
notation to help
us and
generalise the
coefficients with
a formula based
on what was
observed here.

5. !
!!
!The Factorial Symbol
0! = 1 1! = 1
n! = n(n-1) · . . . · 3 · 2 · 1
n must be an integer greater than or equal to 2
What this says is if you have a positive integer followed by
the factorial symbol you multiply the integer by each integer
less than it until you get down to 1.
6! = 6 · 5 · 4 · 3 · 2 · 1 = 720
Your calculator can compute factorials. The !
symbol is under the "math" menu and then "prob".

6. ( )
If and are integers with 0 ,
the symbol is defined as
!
! !
j n j n
n
j
n n
j j n j
≤ ≤
 
 ÷
 
 
= ÷
− 
This symbol is read "n taken j at a time"
Your calculator can compute these as well. It is also under
the "math" and then "prob" menu and is usually denoted nCr
with the C meaning combinations. In probability, there are n
things to choose from and you are choosing j of them for
various combinations.

7. Let's work a couple of these:
( )
5 4 3 2 1
2 1 3 2 1
× × × ×
=
× × ×
20
10
2
= =
( )
12 12!
9 9! 12 9 !
 
= ÷
−  ( )
12 11 10 9!
9! 3 2 1
× × ×
=
× ×
220=
We are now ready to see how this applies to expanding
binomials.
2
( )
5 5!
2 2! 5 2 !
 
= ÷
− 
( )
!
! !
n n
j j n j
 
= ÷
− 

8. 4 8
495a x=
( ) 1
0 1
n n n nn n n
x a x ax a
n
−     
+ = + + + ÷  ÷  ÷
     
L
The Binomial Theorem
The x's start out to the nth power and decrease by 1 in power
each term. The a's start out to the 0 power and increase by 1
in power each term. The binomial coefficients are found by
computing the combination symbol. Also the sum of the
powers on a and x is n.
Find the 5th term of (x + a)12
5th term will have a4
(power on a is 1 less than term number)
So we'll have x8
(sum of two powers is 12)
4 812
4
a x
 
 ÷
 1 less
than
term
number

9. Here is the expansion of (x + a)12
…and the 5th term matches the term we obtained!
In this expansion, observe the following:
•Powers on a and x add up to power on binomial
•a's increase in power as x's decrease in power from
term to term.
•Powers on a are one less than the term number
•Symmetry of coefficients (i.e. 2nd term and 2nd to last term
have same coefficients, 3rd & 3rd to last etc.) so once you've
reached the middle, you can copy by symmetry rather
than compute coefficients.

10. Let's use what we've learned to expand (2x - 3y)6
First let's write out the expansion of the general (x + a)6
and
then we'll substitute.
( )
6 6 5 2 4 3 3 4 2 5 6
__ __ __ __ __x a x ax a x a x a x a x a+ = + + + + + +
these will be the same
these will be the same
Let's find the
coefficient for the
second term.
6 6! 6 5!
6
1 1!5! 5!
  ×
= = = ÷
 
Let's confirm that
this is also the
coefficient of the
2nd to last term.
6 6! 6 5!
6
5 5!1! 5!
  ×
= = = ÷
 
6 6
Let's find the
coefficient for the
third term.
6 6! 6 5 4!
15
2 2!4! 2 4!
  × ×
= = = ÷
× 
15
This will
also be the
coefficient of
the 3rd to
last term.
15
Now we'll find
the coefficient of
the 4th term
6 6! 6 5 4 3!
20
3 3!3! 3 2 3!
  × × ×
= = = ÷
× × 
20
Now we'll apply this formula to our specific binomial.
Instead of x
we have 2x
Instead of a
we have -3y
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
6 6 5 2 4
3 3 4 2 5 6
6 15
20
2 2 2 2
2 215
3 3 3
3 3 3 36 2
y y y
y y y
x x x
x x y
x
x
= + + +
+ + +
− − −
− − − −
6 5 4 2 3 3
2 4 5 6
64 576 2160 4320
4860 2916 729
x x y x y x y
x y xy y
= − + −
+ − +

11. Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au