You first learned how to multiply bimomial expressions in Grade 10. They looked like this: (x - 4)(x + 5)
You used a method called 'F.O.I.L.'. It also worked when the two binomial expressions were the same:
(x + 5)2 = (x + 5)(x + 5) = x2 + 10x + 25

Now what we'd like to look at is how to work out a binomial expansion for any exponent. For example, how to do a question like this:  (x + 5)5 = (x + 5)(x + 5)(x + 5)(x + 5)(x + 5)

Clearly this would be a lot of work doing it the long way. To convince you how much work it is, we'll attempt it here:
   (x + 5)(x + 5)(x + 5)(x + 5)(x + 5)
= (x + 5)(x + 5)(x + 5)(x2 + 10x + 25)
= (x + 5)(x + 5)(x3 + 15x2 + 75x + 125)
= (x + 5)(x4 + 20x3 + 150x2 + 500x + 625)
= x5 + 25x4 + 250x3 + 1250x2 + 3125x + 3125

Of course, we've left out all the intermediate steps where we multiplied and simplified. The total question took about ten minutes and a full sheet of paper. There has to be an easier way, right?

There is, of course, and it uses a special triangle discovered by a mathematician named Pascal, called appropriately 'Pascal's Triangle'. One of its uses is to work out binomial expansions like the one above.
When we're done, you'll be able to do a problem like this: (2a3 - 5b2)4 in just a few lines.

In order to make the examples easier to explain, we're going to work with a binomial that will contain only variables. The expression we'll use will be (x + y)

Here are the first few expansions of (x + y)n
(x + y)0 = 1
(x + y)1 = 1x + 1y
(x + y)2 = 1x2 + 2xy + 1y2
(x + y)3 = 1x3 + 3x2y + 3xy2 + 1y3
(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4


Look at the last line, for power n = 4. Examine the pattern carefully. First notice how the variables appear in the answer. The first term has the fourth power of x , and no y.
Each term thereafter has one less power of x, and one more power of y.
It ends with the last term, which has no x and y4.

Here's the next power ...
(x + y)5 = 1x5y0 + 5x4y1 + 10x3y2 + 10x2y3 + 5x1y4 + 1x0y5

The powers of x in the answer descend from 5 down to 0.
The powers of y ascend from 0 up to 5.
(Ordinarily you wouldn't bother showing x0 or y0. We show them here just to make the pattern clearer).

Now let's look at the pattern of the coeffients.
Here are the first few expansions again:
(x + y)0 = 1
(x + y)1 = 1x + 1y
(x + y)2 = 1x2 + 2xy + 1y2
(x + y)3 = 1x3 + 3x2y + 3xy2 +1y3
(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4


The coeffiecients (in red) form a distinct pattern. The pattern will become more obvious when we write them again as a symmetrical triangle:
Each row in the triangle begins and ends with a 1
To get the other numbers, you add the two numbers in the row directly above.
Do you see the pattern? The next row will begin and end with a 1.
After the first 1, the next number will be the sum of 1 and 4, which is 5,
and so on .... here's the triangle with the next row added:

You can extend Pascal's Triangle easily to as many rows as needed. To do a binomial expansion, you need to know that the rows correspond to the power, beginning with 0.

So (x + y)0 = 1  and  (x + y)1 = 1x + 1y  and  (x + y)2 = 1x2y0 + 2x1y1 + 1x0y2

and so on.

If we wanted the answer for (x + y)4, we'd use the coeffiecients  1, 4, 6, 4, 1 and the pattern for powers of x and y, and we'd get:
(x + y)4 = 1x4y0 + 4x3y1 + 6x2y2 + 4x1y3 + 1x0y4


You don't even need to write out Pascal's Triangle every time you want to do an expansion. If you've studied combinations in Math 30, you can use them to generate any row of the triangle.
Here's an example. We'll do the next row.

(x + y)5
= 5C0x5y0 + 5C1x4y1 + 5C2x3y2 + 5C3x2y3 + 5C4x1y4 + 5C5x0y5


Your calculator will work out the 'choose' values if you don't remember them.
So the expansion can be done in just one line:

(x + y)5 = 1x5y0 + 5x4y1 + 10x3y2 + 10x2y3 + 5x1y4 + 1x0y5

Here's the next row, using 'choose' notation:

(x + y)6 = 6C0x6 + 6C1x5y1 + 6C2x4y2 + 6C3x3y3 + 6C4x2y4 + 6C5x1y5 + 6C6y6

Very easy!

We have a small calculator that will do any power expansion automatically for you, and show you the triangle as well. Give it a try here. Very handy! Notice that it doesn't bother showing powers of 0, or exponents 1.

We have something else for you as well. Someone a long time ago discovered that when you show many rows of Pascal's Triangle, and then go through all the rows and colour numbers depending on what they're divisible by, you get some fascinating patterns.


Pascal's Triangle, with all numbers divisible by 6 coloured.

You need to see the patterns to believe them ... they are fractal-like, and each multiple gives a different pattern. Download this little program, called 'Pascal's Triangle', (Just 89k, zipped, and it works in Windows 11) and give it a whirl. You'll be amazed at the patterns it shows. There is probably a lot of math hidden in the patterns, as well. You might think about why, for instance, some numbers give the patterns they do and others don't.

O,K., back to binomial expansions. We now want to be able to do a problem like this:
(2a3 - 5b2)4
where we aren't just dealing with x and y, but other variables with their own exponents.

The method we outlined above makes this problem a lot easier than it would have been otherwise.
Move on to page 2 and we'll show you how it's done.



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