 x  a
n
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.
0
x  a 1
Can you see a
1
pattern?


 x  a  x  a
2
2
2
 x  a   x  2ax  a
Can you make a
guess what the next
one would be?
3
2
2
3
x

a

x

3
ax

3
a
x

a


3
 x  a   x  4ax  6a x  4a x  a
5
5
4
2 3
3 2
4
5
x

__
ax

__
a
x

__
a
x

__
a
x

a
x

a



4
4
3
2
2
3
4
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.
Let's list all of the coefficients on the x's and the a's and look for a
pattern.
 x  a
1
5
 1x5  5ax 4  10a 2 x3  10a 3 x 2  5a 4 x  1a 5
1+1
1+2+1
 x  a  1
1
 x  a   1x  1a
2
2
2
x

a

1
x

2
ax

1
a


0
1+3 3 1
+ +
1 + 4 + 6 + 4 +1
1
5
3
2
2
3
x

a

1
x

3
ax

3
a
x

1
a


3
 x  a
4
10 10
 1x  4ax  6a x  4a x  1a
4
3
2
2
3
4
5
1
Can you
guess the
next row?
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.
1
1
1
1
1
1
2
3
4
5
1
1
3
6
1
4
10 10
1
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.
!
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".
!
If j and n are integers with 0  j  n,
n
the symbol   is defined as
 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.
n
n!
 
 j  j ! n  j  !
Let's work a couple of these:
5
5!
5  4  3  2 1
20


 10
 
2
 2  2! 5  2  ! 2 1 3  2 1
2
12 
12!
12 1110  9!

 220
 
 9  9!12  9 ! 9! 3  2 1
We are now ready to see how this applies to expanding
binomials.
The Binomial Theorem
 x  a
n
 n  n  n  n 1
   x    ax 
0
1
n n
 a
 n
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.
12  4 8
4
Find the 5th term of (x + a)12
5th term will have a4
(power on a is 1 less than term number)
1 less
than
term
number
8
a
x

495a
x
 
4
So we'll have x8
(sum of two powers is 12)
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.
Instead of x
Instead of a
we have 2x
we have -3y
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.
 x  a
6
6 ax 5  15
20 a 3 x 3  15
6 a5 x  a6
 x 6  __
__ a 2 x 4  __
__ a 4 x 2  __
3 y will
2 xbe
 2 x  3 y    2 x   6  these
 the15same
 3 y   2 x  
3
3
4
2
5
6
will
20  3 y   2 x   15  3these
y 2
x bethe
6 same
3 y   2 x    3 y 
6
6
5
2
4
Let's confirm
that
Let's
find
the
This
will
6


6! 6  5!
Let's
find
the
also3be
the
Now
we'll find

6
6 the 6
5 6!6!  6 655
44
2 this is also
3the
6
coefficient
for





4!
3!

coefficient
forofthe

64
x

576
x
y

2160
x
y

4320
x
y
1
1!5!
5!






15
20
coefficient
of
the of
the
coefficient


second term.
coefficient




third
3 2 3!3!
2!4! 5 3 224!
3! 62nd tothe
last3rd
term.
the
4thterm.
term
2 4   
to
4860 x y  62916
729 y
6! xy 6 5!
last term.



6
 formula

Now we'll apply this
to our 5!
specific binomial.
 5  5!1!
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