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Solving (2x+3)^4


Date: 01/22/2001 at 04:51:30
From: Ali Sadreddini
Subject: Mathematics, Algebra

I think I have found a new way to solve a problem easily, quickly, and 
in only one step. 

   (2x+3)^4

My answer is this:

   (2x)^(4)+12*(2x)^(3)+54*(2x)^(2)+108*2x+81

but could you tell me how to find it, and how many steps it takes to 
answer this?

Regards.


Date: 01/22/2001 at 12:35:33
From: Doctor Peterson
Subject: Re: Mathematics, Algebra

Hi, Ali.

There's a pretty quick way to do this. I simply remember a row from 
Pascal's Triangle, 1 4 6 4 1, and write down this fact:

    (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

Your result follows easily, by replacing a with 2x and b with 3.

You can read about this in the Dr. Math FAQ on Pascal's Triangle:

    http://mathforum.org/dr.math/faq/faq.pascal.triangle.html   

If your method is different from this, I'd like to hear what it is. 
Even if it's not as fast, it would be interesting to see what sort of 
thinking you've done.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   


Date: 01/23/2001 at 04:19:51
From: Ali Sadreddini
Subject: Re: Mathematics, Algebra

Hello, and thank you very much for your fast response. But I wasn't 
very happy when I heard your answer, because I thought I might be the 
first person to solve this kind of expression the way I do it. 

I will explain my way to you:

When we have (2x+3)^(6), my answer is done this way:

   (2x)^(6)+18*(2x)^(5)+135*(2x)^(4)+540*(2x)^(3)+1215*(2x)^(2)
     +1458*2x+729

1. When we have 2x in the beginning, I put a power of 6 to the x.
2. Then I find 18, the product of 6 and 3.
3. This is the most important step of the answer: we divide 18 
   by 2, then multiply it by the power, which is 5, and then multiply
   it by 3 according to (2x+3).
4. This step shows the same way of finding the number except we have  
   to replace division of 2 by 3.
5. Then we continue the calculations until we get to a number without 
   any x, or power.

In this expression that number is 729.

But the final answer is done when you multiply the numbers we have 
found by 2, according to (2x+3).

If you already knew this way, could you please tell me so I can stop 
my research, because I don't want to waste my time on something that 
has been done before.

By the way, i am 15 years old, from Jamison High School, Penrith, 
NSW, Australia, but I am originally from Iran. Thanks for your time. 

Ali Sadreddini


Date: 01/23/2001 at 13:17:43
From: Doctor Peterson
Subject: Re: Mathematics, Algebra

Hi, Ali.

Don't feel at all bad that what you discovered is already known; it 
wasn't known to you, so you've simply done what great mathematicians 
have done in the past. It's not likely that you will find something 
entirely new at your age, but that you found something like this is 
great! I enjoyed this kind of experimentation at your age, too, and 
though little of what I discovered was really very significant, it was 
good practice for the future.

The method you are using is essentially the standard way (which I left 
out) to construct a row of Pascal's triangle: start with 1, then 
multiply first by 4/1 (giving 4), then by 3/2 (giving 6), then by 2/3 
(giving 4), then by 1/4 (giving 1), always decreasing the numerator 
and decreasing the denominator. You are combining this with the 
appropriate factors from the "a" and "b" terms 2x and 3, by starting 
with 2^4 and multiplying also by 3/2 at each step. That's a nice 
addition to the method, and does indeed make it a bit faster.

I think you'll enjoy the study of Pascal's triangle, because there are 
a lot of similar tricks, and some very surprising ways to use it in 
addition to binomials. Have fun with it, and let us know of any 
further discoveries so we can share the adventure.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   


Date: 01/29/2001 at 04:43:41
From: Ali Sadreddini
Subject: Re: Mathematics, Algebra

Hello again.

I wanted to ask you something about Pascal's triangle: how do you find 
the answer for an algebraic expression like this:

   (2x+3)^4

Do you have to continue the triangle and then find the numbers, or is 
there another way?  How many steps does it take?

Ali


Date: 01/29/2001 at 09:42:07
From: Doctor Peterson
Subject: Re: Mathematics, Algebra

Hi, Ali.

I'm not sure what you are asking; you don't need to continue the 
triangle beyond the rows that are usually shown. I'll just repeat what 
I said before, with a more complete explanation of the relation 
between the triangle and your method. Let me know if I have missed 
your question.

As you see in our FAQ on Pascal's triangle,

   http://mathforum.org/dr.math/faq/faq.pascal.triangle.html   

the expansion of

   (x+1)^4 = 1 + 4x + 6x^2+ 4x^3 + x^4

has coefficients taken from the fourth row of the triangle, 1 4 6 4 1 
(counting the top row as row zero). For the more general expansion of 
(a+b)^n, this becomes

   (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

This can be done just by knowing the first few rows of the triangle, 
as I did initially; but for larger powers it is not necessary to 
construct the whole triangle to find the row you need.

The formula for the nth row of the triangle, again from the FAQ, is

              n!        n(n-1)...(k+1)
   [n:k] = -------- = ------------------
           k!(n-k)!   (n-k)(n-k-1)...(1)

For example,

            4*3*2*1    4*3
   [4:2] = --------- = --- = 6
           2*1 * 2*1   2*1

The expansion can then be rewritten as

   (a+b)^4 = [4:0]a^4 + [4:1]a^3b + [4:2]a^2b^2 + [4:3]ab^3 + [4:4]b^4

Your method, which is better than just memorizing the triangle, 
depends on the recursion formula:

   [n:k+1] = (n-k)/(k+1) * [n:k]

which we can demonstrate this way:

                    n!                 n!             n-k      n!
    [n:k+1] = -------------- = -------------------- = --- * --------
              (k+1)!(n-k-1)!   (k+1)k! (n-k)!/(n-k)   k+1   k!(n-k)!

For example, for n = 4 and k = 2,

    [4:3] = 2/3 * [4:2] = 2/3 * 6 = 4

Putting this all together, starting with the first term, a^4, we can 
get the successive terms by multiplying by (n-k)/(k+1) to get the 
coefficient, and by b/a to get the proper powers of a and b:

    n  k  term     multiply by
    -  -  ------   -----------
    4  0  a^4      4/1 b/a
    4  1  4a^3b    3/2 b/a
    4  2  6a^2b^2  2/3 b/a
    4  3  4ab^3    1/4 b/a
    4  4  b^3      0

This, as I understand it, is your method, which is better than 
directly referring to Pascal's triangle. You just multiply by a 
fraction whose numerator decreases while the denominator increases, 
until you reach zero.

I haven't run across anything in our archives that gives this 
recursive formula, though it's well-known. Again, you're to be 
commended for finding it! Let me know if you have any more questions.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Polynomials

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