f(x) be a polynomial of degree atmost 7 which leaves remainder -1 and 1 on division by \((x-1)^4\) and \((x+1)^4\) respectively. If the sum of pairwise product of all roots of \(f(x)=0\) is n then the value of (5n + 24) is

P.S. final answer is 0, but can someone elaborate the solution?

Thank you! :)

amygdaleon305 Jul 28, 2021

#1**+2 **

I actually do not understand what this means

"If the sum of pairwise product of all roots of f(x)=0 is n" Is it even meant to be real like that??

I thought it meant that if you find the product of every pair of roots and add them together then the answer is 0.

But that would mean that n has no meaning in the problem.

So I am really confused.

Amy, do you understand what the question is asking? If so please explain it to me.

Melody Aug 1, 2021

#2**+2 **

Melody,

From what I understand it says that f(x) is a polynomial of degree *atmost 7* right? then it would have *atmost 7 roots*.

Let's then consider its roots are a,b,c,d,e,f,g then, *sum of pairwise product* of all roots means **ab + bc + cd + de + ef + fg + ag + bg + cg +...+ eg + af + bf +.... so on = n** you get the point right?

I think then we would've to find that sum to find the value of n.

Another thing, its also given that f(x) leaves remainder -1 and 1 on division with (x-1)^{4} and (x+1)^{4} I don't know what to deduce from that...

amygdaleon305
Aug 1, 2021

#3**+2 **

Woah, this is a hard problem, I can't figure it out, but maybe this is helpful.

f(x) = a(x-1)^4 - 1, where a is at most 3 degrees.

f(x) = b(x+1)^4 + 1, where b is at most 3 degrees.

This would mean that f(1) = -1, f(-1) = 1.

The sum of the coeficients is -1, and the even degree coefficients - odd degree coefficients = 1.

By vieta's, ax^(n) + bx^(n-1) + cx^(n-2)..., the pairwise product of all the roots would be c/a.

Good luck. :))

=^._.^=

catmg Aug 1, 2021

#4**+2 **

Yeah that's very good but then in order to find the value of n we would've to find the value of c/a which are the coefficients in f(x).

Anyway, thanks a lot catmg! :D Let me see what can be done further

amygdaleon305
Aug 1, 2021

#5**+1 **

Has anyone had any luck with this question so far? I tried everything I could, but it only seem to get more confusing

amygdaleon305 Aug 2, 2021

#7**+1 **

Its okay, its a really hard question... it may take some time to get solved

amygdaleon305
Aug 4, 2021