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The Easy Peasy Binomial Expansion Trick

October 15th, 2012 by Math Tricks | 5 Comments | Filed in Math Tricks


In a previous post, I showed you how to use Pascal’s triangle to quickly expand binomial equations.  There is another math trick that will help you to expand binomials fast.  No – it is not the now infamous joke that many of you have now seen – i.e.:



(4x + 16y)7



(4x  +  16y)7


(4x   +   16y)7


(4x    +    16y)7


(4x     +     16y)7


(4x      +      16y)7


(4x       +       16y)7



Funny, funny!  But seriously, if you are in need to quickly expand a binomial, and you do not have a detailed Pascal’s triangle readily available, this binomial expansion trick will work out very well for you.


First, let’s set up as our example problem:


(x – y)5


To expand, first start with the x’s.  Begin by writing xN (where N is the exponent the x + y term is being raised to), and continue to xN-1, xN-2 … xN-N:


x5               x4               x3               x2               x1               x0


Note that x0 is simply equal to 1.


Now do the same for the y terms, but in reverse order:


x5y0              x4y1              x3y2              x2y3              x1y4              y5


Again, note that y0 is equal to 1.


The next step is to write in the “+” or “-“ operators.  Since the operation within the parentheses is a subtraction, alternate the operators, beginning with the “+” operator.  Here, the x5y0 is positive, and our example becomes:



x5     –      x4y1     +      x3y2     –       x2y3     +      x1y4     –      y5


Note: if the terms within the parentheses were added (i.e., x + y), then only the “+” operator is used for all terms.


Now, write down the term number of each term below each term.  For small exponents, this may not be necessary for you to do, but for large expansions, it is best to do this so that you do not lose track of your term numbers:


x5     –      x4y1     +      x3y2     –       x2y3     +      x1y4     –      y5

1               2                 3                   4                  5               6


So now comes the part where the coefficients for each term are written.  This is very easy to do with the way we set up our example.


Firstly, the coefficients for the first term and the last term are always 1.  For the coefficients of the other terms, simply take the coefficient of the previous term, multiply that by the “x” exponent of the previous term, and then divide the result by the term number of the previous term.  So in our example, the coefficient of the second term would be:


(1 x 5)/1 = 5


Now write down the coefficient and continue on:


x5     –      5x4y1     +      x3y2     –       x2y3     +      x1y4     –      y5

1               2                   3                   4                  5               6



The coefficient of the third term is:


(5 x 4)/2 = 10


x5     –      5x4y1     +      10x3y2     –       x2y3     +      x1y4     –      y5

1                2                      3                   4                  5                6



The coefficient of the fourth term is:


(10 x 3)/3 = 10


x5     –      5x4y1     +      10x3y2     –       10x2y3     +      x1y4     –      y5

1                 2                      3                       4                  5               6



The coefficient of the fifth term is:


(10 x 2)/4 = 5


x5     –      5x4y1     +      10x3y2     –       10x2y3     +      5x1y4     –      y5

1                 2                      3                       4                   5                6


Again, the coefficient of the last term is always 1, so our binomial expands out to:


x5 – 5x4y1 + 10x3y2 – 10x2y3 + 5x1y4 – y5   answer



Not so bad, eh?  Good!  Now you can practice this technique a couple of times and you will soon be very efficient at it.  Remember, for more complicated terms, you can use substitution while applying this technique, and then plug back in after you expand the binomial (e.g., if your terms within the parentheses were “4.2w + 33z”,  set x=4.2w and y=33z, and expand it as “x + y”).


Below are a couple of binomials together with their expansions.  Try to apply this technique to these binomials and see if your answer agrees with the answers given.




(x + y)4



x4 + 4x3y + 6x2y2 + 4xy3 + y4




(4x + 16y)7



(4x)7 + 7(4x)6(16y) + 21(4x)5(16y)2 + 35(4x)4(16y)3

+ 35(4x)3(16y)4 + 21(4x)2(16y)5 +7(4x)(16y)6 + (16y)7




(x – y)9



x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4 – 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9







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The Binomial Theorem and Pascal’s Triangle

February 24th, 2010 by Math Tricks | 20 Comments | Filed in Math Patterns, Math Tricks, Pascal's Triangle

Back in grade school, I was first introduced to the Binomial Theorem.  The title alone was quite enough to intimidate me, let alone the seemingly impossible to understand equations involved with it.

I’ll not go into the mathematics of the binomial theorem here.  Instead, I’ll introduce you to math tricks which can be used instead.  First, let me refresh your mind on why we were taught the binomial theorem.  Remember when you were asked to expand the equation:

(x + y)2

If you recall, this equation can be expanded to the equivalent equation:

x2 + 2xy + y2

The binomial theorem will allow you to solve a higher order problem of the example above.  For instance, what is the expansion of the equation:

(x + y)5

Generally, an equation of this type can be expanded as:

binomial expansion

where c1, c2, … are the binomial coefficients in the expansion.   So given any n, you can determine the expansion without the coefficients.  Expanding our example above:

(x + y)5 = c1x5y0 + c2x4y1 + c3x3y2 + c4x2y3 + c5x1y4 + c6x0y5

So how do you determine the binomial coefficients?  You can determine the binomial coefficients individually using the equation:

binomial coefficients

for k=0 to k=n.  This works fine, but is a little bit cumbersome – especially for large values of n!  So what is the math trick to solve this quickly?

Before I can answer this, I have to introduce to you Pascal’s Triangle.  Pascal’s triangle is a mathematical progression which is determined by constructing a triangle with numbers using a very simple algorithm.  First, take a look at this example of Pascal’s triangle:

hexagonal pascal triangle

At the very top is row 0, which is simply a 1.  In row 1, there are two numbers, both 1s.  In row 2, there are three numbers: 1, 2, and 1.  Notice that the 2 in row two is the sum of the two numbers above it; this is how you determine the numbers in the triangle – simply add two side-by-side numbers to get the result below and between the numbers:

Animated Pascal TriangleConstruction of Pascal’s Triangle1

So how can you use Pascal’s triangle to find the binomial coefficients when you expand the equation (x + y)5?  First, notice that the equation is raised to the 5th power.  So now simply go to the 5th row of Pascal’s triangle (remember, the top row is row 0), and those numbers are the required coefficients:

1 5 10 10 5 1

And so,

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

Quite a time saver!


For an alternative method of expanding polynomials, please check out The Easy Peasy Binomial Expansion Trick.

1File by Hersfold, en.wikipedia.org/wiki/User:Hersfold

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