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Looking for Patterns


Date: 10/30/2001 at 17:47:00
From: Amanda
Subject: (no subject)

What would be the answer to: (x-a)(x-b)(x-c)(x-d)(x-e)...(x-y)(x-z)? 

Thanks.


Date: 10/31/2001 at 02:07:23
From: Doctor Jeremiah
Subject: Re: (no subject)

Hi Amanda,

It depends. Do we know the values of any of these letters? Are any of 
them multiples of any of the others? If they are all different, and 
are not multiples of each other, then we can look for a pattern by 
starting small and working our way up:

           (x-a)(x-b) = x^2
                      - ax - bx
                      + ab

      (x-a)(x-b)(x-c) = x^3
                      - cx^2 - ax^2 - bx^2
                      + cax + bcx + abx
                      - abc

 (x-a)(x-b)(x-c)(x-d) = x^4
                      - dx^3 - cx^3 - ax^3 + dcx^2 + adx^2 - bx^3
                      + dbx^2 + cax^2 + bcx^2 + abx^2
                      - dcax - dbcx - dabx - abcx
                      + dabc

See the pattern? For 22 terms the first line would be x^22, and then 
every combination of 1 letter (26 of them) times x^21 would be 
subtracted, and then every combination of 2 letters times x^20 would 
be added, and then every combination of 3 letters times x^19 would be 
subtracted, and so on and so on until in the last line every 
combination of 26 letters (1 of them) times x^0 (which is 1) would be 
added.

Try writing it down. It will be enormous! That's why it's always 
easier to look for patterns and extrapolate.

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


Date: 11/14/2001 at 20:47:23
From: Natalie
Subject: Value of an expression

What is the value of the following expression:

(x-a)(x-b)(x-c)...(x-z)

so that there are a total of 26 factors, with each letter of the 
alphabet subtracted from x in one of the factors?

I have tried to answer it and so have my parents and friends, but we 
are all confused.

Please help!


Date: 11/15/2001 at 16:19:56
From: Doctor Greenie
Subject: Re: Value of an expression

Hi, Natalie -

I hope this is not a trick question.  It is possible that the answer
is 0, because one of the factors is (x-x), which is 0, and 0
multiplied by a bunch of other factors is still 0.

I will assume it is not supposed to be a trick question, because it is
kind of fun to try to imagine what the product of 26 binomials would
look like.

I don't know how you are going to write out the value of this
expression. If you wrote it all out explicitly (showing every term),
it would take probably a few dozen pages. Let's look at similar cases
with smaller numbers of factors and see what those expressions look
like in expanded form.

Let's look first at the case with two factors....

  (x-a)(x-b) = x^2 - ax - bx + ab

This result comes from your familiar FOIL method of multiplying
binomials.

The FOIL method of multiplying binomials is a simplification of a
general algorithm for multiplying polynomials. The general algorithm
says that the product of any number of polynomials is the sum of all
the possible "partial products" obtained by selecting one of the terms
from each polynomial. For example, with the FOIL method for
multiplying two binomials, we have

  F ("first") = first term * first term
  O ("outer") = first term * second term
  I ("inner") = second term * first term
  L ("last")  = second term * second term

Where does the x^2 term come from in this product? It comes from
selecting the x term in each of the two factors.

Where do the two x terms come from?  They come from selecting the x
term from one of the two factors and the constant term (-a or -b) from
the other factor.

And where does the constant term come from?  It comes from selecting
the constant terms from both factors.

So let's rewrite this product as

  (x-a)(x-b) = x^2 - (a+b)x + ab

In this expression, we have...

(1) an x^2 term with coefficient 1;
(2) an x term with a coefficient which is the sum of the constants in 
    the two factors; and
(3) a constant term equal to the product of the constants in the two 
    factors

Now let's look at the case with three factors.

  (x-a)(x-b)(x-c) = ?

Where will the x^3 term(s) come from in the product?  From selecting
the x term from each of the three factors. This gives us a term x^3 
in the product.

Where will the x^2 terms come from in the product? From selecting the
x term from two of the three factors and the constant from the other
term. These terms give us -(a+b+c)x^2 in the product.

Where will the x terms come from in the product? From selecting the 
x term from one of the three factors and the constant term from the
other two. All the constant terms are negative; if I select two of
them, then their product is positive. And selecting the constant
factors in two of the three terms gives me the possible combinations
ab, ac, and bc. So these terms give me +(ab+ac+bc)x in the product.

And the constant term comes from selecting the constant term in each
of the three factors. This gives me a term -abc in the product.

So this product will be

  (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc

And let's look at one more case, with four factors....

  (x-a)(x-b)(x-c)(x-d) = ?

Where will the x^4 term(s) come from in the product?  From selecting
the x term from each of the four factors. This gives us a term x^4 in
the product.

Where will the x^3 terms come from in the product? From selecting the
x term from three of the four factors and the constant from the other
term. These terms give us -(a+b+c+d)x^3 in the product.

Where will the x^2 terms come from in the product? From selecting 
the x term from two of the four factors and the constant term from the
other two. All the constant terms are negative; if I select two of
them, then their product is positive. And selecting the constant
factors in two of the four terms gives me the possible combinations
ab, ac, ad, bc, bd, and cd. So these terms give me 
+(ab+ac+ad+bc+bd+cd)x^2 in the product.

Where will the x terms come from in the product?  From selecting the 
x term from 1 of the 4 factors and the constant term from the other
three. All the constant terms are negative; if I select three of them,
then their product is negative. And selecting the constant factors in
three of the four terms gives me the possible combinations abc, abd,
acd, and bcd. So these terms give me -(abc+abd+acd+bcd)x in the
product.

And the constant term comes from selecting the constant term in each
of the 4 factors. This gives me a term abcd in the product.

So this product will be

  (x-a)(x-b)(x-c)(x-d) = x^4 - (a+b+c+d)x^3 + (ab+ac+ad+bc+bd+cd)x^2
                         - (abc+abd+acd+bcd)x + abcd

Now let's generalize the pattern.  In the product of n factors

  (x-a)(x-b)(x-c)(x-d)(x-e)...

we will get

(1) terms of x^n, x^(n-1), x^(n-2), ..., x^2, x, and a constant term;

(2) the terms in decreasing powers of x will have the x^n term 
    positive, with the coefficients of successive terms having 
    alternating signs;

(3) the x^n term will come from selecting the x term from all n 
    factors; there is only one way to do this, so the coefficient of 
    the x^n term is 1;

(4) the x^(n-1) term will come from selecting the x term from n-1 of 
    the factors and the constant term from the other, so the 
    coefficient of the x^(n-1) term will be -(sum of all the 
    "products" of the constant terms taken 1 at a time);

(5) the x^(n-2) term will come from selecting the x term from n-2 of 
    the factors and the consant term from the other 2, so the 
    coefficient of the x^(n-2) term will be +(sum of all the 
    "products" of the constant terms taken 2 at a time);
...

(?) the x^2 term will come from selecting the x term from two of the 
    factors and the consant term from the other (n-2) terms, so the 
    coefficient of the x^2 term will be (+ or -)(sum of all the 
    "products" of the constant terms taken (n-2) at a time);

(?) the x term will come from selecting the x term from 1 of the
    factors and the consant term from the other (n-1) terms, so the 
    coefficient of the x term will be (+ or -)(sum of all the 
    "products" of the constant terms taken (n-1) at a time);

(?) the constant term will come from selecting the constant term in 
    each factor, so the coefficient will be (+ or -)(product of all of 
    the n constant terms)

So the "answer" to your question is the following:

  (x-a)(x-b)(x-c)...(x-y)(x-z) =

   x^26
 - (sum of all "products" of constant terms taken 1 at a time)x^25
 + (sum of all products of constant terms taken 2 at a time)x^24
 - (sum of all products of constant terms taken 3 at a time)x^23
 + (sum of all products of constant terms taken 4 at a time)x^22
 ...
 + or - ...
 ...
 + (sum of all products of constant terms taken 24 at a time)x^2
 - (sum of all products of constant terms taken 25 at a time)x
 + (product of constant terms taken 26 at a time)

The total number of terms in the expansion is 2^26 = 67108864 (because 
there are two choices for the term to select in each of the 26
factors), so it is clearly unreasonable to write out the expansion
explicitly. 

I hope this helps.  Write back if you have any further questions on
this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Sequences, Series

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