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| Topic: can you solve this algebra?? | |
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Edited by
prithvii1989
on
Wed 05/06/15 12:08 AM
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(x-a)(x-b)(x-c)(x-d).....(x-y)(x-z) = ??
please write the answers only, dnt disclose how u get it. will disclose the right answer later. |
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0
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Edited by
debbie1980
on
Wed 05/06/15 01:20 AM
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the answer is 0
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X=[$#!+]
Really, is this even approachable without giving value to the variable? Its been a while but there should be at least 26 values before this can be solved. |
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I think its a trick question
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Edited by
Amelinng
on
Wed 05/06/15 03:38 AM
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0 ....
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yup it deffo 0 well done prithi
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Edited by
prithvii1989
on
Wed 05/06/15 03:49 AM
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0 .... i wrote not to explain your answers. Million apologies! |
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0 .... i wrote not to explain your answers. Million apologies!  ...i guess that ends this chit chat
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0 .... i wrote not to explain your answers. Million apologies! ...i guess that ends this chit chat
Don't rub it in.... |
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0 .... i wrote not to explain your answers. Million apologies! its ok. it happens sometimes. |
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0 .... i wrote not to explain your answers. Million apologies! ...i guess that ends this chit chat
Don't rub it in.... 
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what you people are trying to rub in.
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We have to find the sums of the products in (X-a)(X-b)(X-c)....(X-y)(X-z);where X is a variable and a, b, c, d,..z are constants.
Consider; (X-a)(X-b) = X2 - aX - bX + ab = X2 -(a+b)X + ab Two binomials (or 2 factors) give 3 sum of product terms. (X-a)(X-b)(X-c) = X3 - aX2 - bX2 -cX2 + abX + acX + bcX - abc = X3 - (a+b+c)X2 + (ab+ac+bc)X - abc Three binomials (or 3 factors) give 4 sum of product terms. Here, 26 factors in (X-a)(X-b)(X-c)....(X-y)(X-z). (X-a)(X-b)(X-c)...(X-y)(X-z) = X26 - (sum of all products of constant terms taken 1 at a time)X25 + (sum of all products of constant terms taken 2 at a time)X24 - (sum of all products of constant terms taken 3 at a time)X23 + (sum of all products of constant terms taken 4 at a time)X22 ... + or - ..... + (sum of all products of constant terms taken 24 at a time)X2 - (sum of all products of constant terms taken 25 at a time)X + (product of constant terms taken 26 at a time) Hence, the required number of terms is 27. |
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0 .... i wrote not to explain your answers. Million apologies! ...i guess that ends this chit chat
Don't rub it in....
Phew..... what a relief! Sorted that out......!!! what you people are trying to rub in.
LOL.......!! 
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(x-a)(x-b)(x-c)(x-d).....(x-y)(x-z) = ?? please write the answers only, dnt disclose how u get it. will disclose the right answer later. Its 0... and how did I get that answer.. the old fashioned way. I scrolled down and got it from someone else...I cheated!.... just like in High school |
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At some point in the series your reach the part of the equation that reads (x-x), which is being multiplied across the rest of the equation.
All who said the answer is 0, get a cookie. |
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wtf...math ....uuuuuugh
 
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wtf...math ....uuuuuugh
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