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# Can't seem to figure this.. (Algebra)

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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! :)

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. 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. :))

=^._.^=

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 Aug 2, 2021
#6
+1

I can't seem to figure it out. :((

=^._.^=

catmg  Aug 3, 2021
#7
+1

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

amygdaleon305  Aug 4, 2021