Let \(g(x)\) be a polynomial such that \(g(x)=x^{8}-4x^{7}+7x^{6}+ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f\) factorizes into \(8\) linear factors.Also roots of the polynomial are all positive,where \(a,b,c,d,e,f\) are all real numbers.Sum of all possible values of \(f\) can be written as \(\frac{a}{b}\),where \(a\) and \(b\) are co-prime positive integers.Find \(a+b?\) I submitted it to brilliant but they rejected,no problem.So you must try it.I want to know multiple methods of this problem.Thank You.After 2 solutions are submitted I will also share my awesome method....!

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TopNewestPerhaps they rejected it because you use \(a,b\) for the coefficients of \(g(x)\) as well as the value of \(f\).

By Vieta's Relations, the sum of the \(8\) roots is \(4\) and the sum of the squares is \(4^2 - 2\cdot 7 = 2\).

Hence, the arithmetic mean of the roots is \(\dfrac{4}{8} = \dfrac{1}{2}\) and the root-mean-square of the roots is \(\sqrt{\dfrac{2}{8}} = \dfrac{1}{2}\).

Therefore, we have the equality case of the RMS-AM inequality. So, all the roots are equal to \(\dfrac{1}{2}\).

Thus, \(g(x) = \left(x - \tfrac{1}{2}\right)^8\) is the unique polynomial \(g(x)\). So, \(f = g(0) = \dfrac{1}{2^8} = \dfrac{1}{256} \leadsto a+b = \boxed{257}\). – Jimmy Kariznov · 3 years, 11 months ago

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– Kishan K · 3 years, 11 months ago

I did the same but I used Cauchy Schwarz Inequality instead of RMS-AM inequality.But nice solution.Log in to reply

Thanks! – Pranav Arora · 3 years, 11 months ago

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Furthermore, \(\sqrt{\dfrac{1}{n}\displaystyle\sum_{k = 1}^{n}x^2_k} = \dfrac{1}{n}\displaystyle\sum_{k=1}^{n}x_k = a\) if and only if \(x_1 = \cdots = x_n = a\).

In this problem, that theorem allows us to conclude that the roots of \(g(x)\) are all equal to \(\tfrac{1}{2}\).

This shows that there is only one possible polynomial \(g(x)\), namely \(g(x) = \left(x - \tfrac{1}{2}\right)^8\). – Jimmy Kariznov · 3 years, 11 months ago

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Mind throwing some light on how you instantly came up with that inequality? – Pranav Arora · 3 years, 11 months ago

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