# The Binomial Theorem

``` 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) &middot; . . . &middot; 3 &middot; 2 &middot; 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 &middot; 5 &middot; 4 &middot; 3 &middot; 2 &middot; 1 = 720
!
Your calculator can compute factorials. The !
symbol is under the &quot;math&quot; menu and then &quot;prob&quot;.
!
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 &quot;n taken j at a time&quot;
Your calculator can compute these as well. It is also under
the &quot;math&quot; and then &quot;prob&quot; 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 &amp; 3rd to last etc.) so once you've
reached the middle, you can copy by symmetry rather
than compute coefficients.
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