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