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tricky algebra problem

Discussion in 'Math' started by chesart1, Nov 24, 2007.

1. chesart1 Thread Starter Senior Member

Jan 23, 2006
269
1
Perhaps this is a good way to teach a student to think before diving into a solution to a problem....

what is the product of the binomials
(x-a)(x-b)(x-c)(x-d)(x-e)(x-f)(x-g)(x-h)(x-i)(x-j)(x-k)......(x-y)(x-z)
??

2. Dave Retired Moderator

Nov 17, 2003
6,960
172
Assuming this sequence in alpha-sequential and there is an (x-x) in there, then the product of the binomials is 0.

Dave

3. recca02 Senior Member

Apr 2, 2007
1,211
1
i must admit i got foxed by this one, since Mr chesart1 wisely made the (x-x) inconspicuous.  4. Dave Retired Moderator

Nov 17, 2003
6,960
172
Yes very sneaky! Here is a question: Removing the (x-x) binomial, how many sum of products are there when the brackets are multiplied out?

Dave

5. recca02 Senior Member

Apr 2, 2007
1,211
1
26. the answer is quite simple isnt it?..but i wont give away any hint now(will edit later perhaps )
i'll be pretty embarrassed if at all i missed the correct answer .

one reason i got tricked was i assumed all of the terms as constants and the 'x' as variable (and was under the impression any other alphabet not shown represent a constant without thinking about the hidden x)

6. Dave Retired Moderator

Nov 17, 2003
6,960
172
Is it 26? The order of the sum of products term will be 26, but will it have 26 product terms. Consider:

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

Two binomials gives 3 sum of product terms.

(x-a)(x-b)(x-c) = x^3 -ax^2 - bx^2 -cx^2 +abx +acx + bcx - abc = x^3 - (a+b+c)x^2 +(ab+ac+bc)x - abc

Three binomials gives 4 sum of product terms.

So I go for 27 - nearly the same!

Whether x is constant, variable, real, complex or other x-x = 0.

Its all fun isn't it! Dave

7. recca02 Senior Member

Apr 2, 2007
1,211
1
yes but i thought one x will be variable and the other constant thats where i went wrong.
for a quadratic we have three terms,
therefore for polynomial of order 25(26-1,since x-x is absent) we should have 26 terms right?....did i go wrong somewhere?
sure is! 8. Dave Retired Moderator

Nov 17, 2003
6,960
172
Ok, I see. Anyone using the same characters for two different variables needs detention!

D'oh! How dumb am I - and I set the question!  Yes 26 is the answer. Dave