# tricky algebra problem

#### chesart1

Joined Jan 23, 2006
269
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)
??

#### Dave

Joined Nov 17, 2003
6,970
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)
??
Assuming this sequence in alpha-sequential and there is an (x-x) in there, then the product of the binomials is 0.

Dave

#### recca02

Joined Apr 2, 2007
1,214
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)
??
i must admit i got foxed by this one, since Mr chesart1 wisely made the (x-x) inconspicuous.  #### Dave

Joined Nov 17, 2003
6,970
i must admit i got foxed by this one, since Mr chesart1 wisely made the (x-x) inconspicuous.  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

#### recca02

Joined Apr 2, 2007
1,214
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
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)

#### Dave

Joined Nov 17, 2003
6,970
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 .
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!

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

Its all fun isn't it! Dave

#### recca02

Joined Apr 2, 2007
1,214
Whether x is constant, variable, real, complex or other x-x = 0.
yes but i thought one x will be variable and the other constant thats where i went wrong.
So I go for 27 - nearly the same!
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?
Its all fun isn't it! sure is! #### Dave

Joined Nov 17, 2003
6,970
yes but i thought one x will be variable and the other constant thats where i went wrong.
Ok, I see. Anyone using the same characters for two different variables needs detention!

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?
D'oh! How dumb am I - and I set the question!  sure is!  