tricky problem in algebra.

[chesart1]

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)
??

 

[erosennin]

welll... it would be x^26 + "don't know" + abcdefghijklmnopqrstuvwxyz
I'll think about this don't know for a bit longer... lol

 

[chesart1]

[hint] I hope nobody tries to do the multiplication. If they do, they are going to be really angry when they find the answer! LOL

 

[chesart1]

welll... it would be x^26 + "don't know" + abcdefghijklmnopqrstuvwxyz
I'll think about this don't know for a bit longer... lol

Sorry! Thats not the answer.

 

[erosennin]

OHHHH I GET IT!!!!!
once you get to (x-x)
!!!!
excelllent!!!!
FANTASTIC

 

[ParkingLotLust]

What's the answer, out of curiosity?

EDIT Clever, now that I think about it. Zero, since (x-x) = 0, which nulls everything else.

 

[vicky877]

Is this rule true or not ?
(x)^0=1
if it is true then justify..

 

[vicky877]

hi

i think it is crazy one question.

 

[Marks256]

what is the answer?! I am going to go with 0 as well.

 

[Salgat]

One of the multiples is (x-x), zero multiplied by the whole thing results in zero.

 

[erosennin]

Is this rule true or not ?
(x)^0=1
if it is true then justify..

x^0 = x^n/x^n = x^(n-n) = 1

 

[RODALCO]

0 or ZERO ! is the total.

Anything multiplied by zero = zero.

 

[Nigel Goodwin]

0 or ZERO ! is the total.

Anything multiplied by zero = zero.

You don't do 'further maths' do you?, apparently it can be any value at all? - and I don't understand it either - but my daughter (doing A level further maths) does!