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### Solving (2x+3)^4

```
Date: 01/22/2001 at 04:51:30
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

(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

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.

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
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.

```

```
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
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

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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