Who can solve this rock solid polynomial?
Simplify the following as much as possible:
(a-x)(b-x)(c-x)(d-x)....(z.
By the way I know the answer - just want to spread the joy and award a lucky person 10 points!
Answers:
There is a term z, therefore there is a term x.
If the variable x and the term x are not differentiated it means that the series contains the term (x-x) which evaluates to zero for all x, therefore the entire expression is equivalent to zero.
You've not got many answers! Not surprising, but I'd be interested to see the answers people give, and in the end, your answer!! How you spread the joy!!
Crikey, curly you are clever
(x-x)=0. Therefore it doesn't matter what everything else is, as it will be multiplied by zero. So the solution is zero.
Curly is right.0
nothing times anything or anything times nothing is nothing, doesn't matter how many other anciliary terms you stick in there, it all counts for nothing!
the answer is zero as (x-x) which is one of the factors is zero
I can't see any error in Curly's reasoning, therefore the whole term regardless of any quantities given must always equal zero.
heavenlyhaggis
You get:
+x^n - Sum($ in alphabet)$ * x^(n-1) + Sum ($,% in alphabet with $<>%) $% * x^(n-2) + . + (-1)^n * Product($ in alphabet)$
where n is the number of letters in your alphabet, and where you need to reverse all the signs if n is odd. This is about the simplest you can get it. Now there are those who would say there is a term (x-x) in the expression, and so therefore, when you simplify it, you get:
0
That's assuming your alphabet has an x in it. Not all alphabets have an x in it. For example, the Italian alphabet does not have an x in it. That dot dot dot (.) hides a lot of detail.
Amongst all the factors, there shall be (x-x) which will equal 0 and 0 times anything will equal 0.
zero
Zero.
(x - x) = 0
Therefore (a-x)(b-x)(c-x)(d-x) . (z-x) = 0
The easy way to solve this is too say (x-x) is eqaul too zero and not matter how many different numbers you muliply by zero you alway get zero.
The hard way to slove this, and expalin fully the mathmatical way, is to mulity out every two sets of brackets.
e.g. (ab-ax-bx+xx)(cd-cx-dx-xx)...
but somehwere along the line you will come too the bracket (wx-wx-xx-xx)
If u you look at this wx-wx is equal too zero, and so is xx-xx.
eventually when you keep mulitpying out the brackets you will end up with xx-xx which as we just found out equals zero.
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(a-x)(b-x)(c-x)(d-x)....(z.
By the way I know the answer - just want to spread the joy and award a lucky person 10 points!
Answers:
There is a term z, therefore there is a term x.
If the variable x and the term x are not differentiated it means that the series contains the term (x-x) which evaluates to zero for all x, therefore the entire expression is equivalent to zero.
You've not got many answers! Not surprising, but I'd be interested to see the answers people give, and in the end, your answer!! How you spread the joy!!
Crikey, curly you are clever
(x-x)=0. Therefore it doesn't matter what everything else is, as it will be multiplied by zero. So the solution is zero.
Curly is right.0
nothing times anything or anything times nothing is nothing, doesn't matter how many other anciliary terms you stick in there, it all counts for nothing!
the answer is zero as (x-x) which is one of the factors is zero
I can't see any error in Curly's reasoning, therefore the whole term regardless of any quantities given must always equal zero.
heavenlyhaggis
You get:
+x^n - Sum($ in alphabet)$ * x^(n-1) + Sum ($,% in alphabet with $<>%) $% * x^(n-2) + . + (-1)^n * Product($ in alphabet)$
where n is the number of letters in your alphabet, and where you need to reverse all the signs if n is odd. This is about the simplest you can get it. Now there are those who would say there is a term (x-x) in the expression, and so therefore, when you simplify it, you get:
0
That's assuming your alphabet has an x in it. Not all alphabets have an x in it. For example, the Italian alphabet does not have an x in it. That dot dot dot (.) hides a lot of detail.
Amongst all the factors, there shall be (x-x) which will equal 0 and 0 times anything will equal 0.
zero
Zero.
(x - x) = 0
Therefore (a-x)(b-x)(c-x)(d-x) . (z-x) = 0
The easy way to solve this is too say (x-x) is eqaul too zero and not matter how many different numbers you muliply by zero you alway get zero.
The hard way to slove this, and expalin fully the mathmatical way, is to mulity out every two sets of brackets.
e.g. (ab-ax-bx+xx)(cd-cx-dx-xx)...
but somehwere along the line you will come too the bracket (wx-wx-xx-xx)
If u you look at this wx-wx is equal too zero, and so is xx-xx.
eventually when you keep mulitpying out the brackets you will end up with xx-xx which as we just found out equals zero.
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