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# Polynomial problem

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....!

Note by Kishan K
4 years, 1 month ago

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Perhaps 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, $$f = g(0) = \dfrac{1}{2^8} = \dfrac{1}{256} \leadsto a+b = \boxed{257}$$. · 4 years, 1 month ago

I did the same but I used Cauchy Schwarz Inequality instead of RMS-AM inequality.But nice solution. · 4 years, 1 month ago

I would like to know why you consider that inequality? I mean you get AM=RMS but where does it really help?

Thanks! · 4 years, 1 month ago

The RMS-AM Inequality states that if $$x_1, \ldots, x_n \ge 0$$, then $$\sqrt{\dfrac{1}{n}\displaystyle\sum_{k = 1}^{n}x^2_k} \ge \dfrac{1}{n}\displaystyle\sum_{k=1}^{n}x_k$$.

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 $$g(x) = \left(x - \tfrac{1}{2}\right)^8$$. · 4 years, 1 month ago

Awesome as always, thanks Jimmy! :)

Mind throwing some light on how you instantly came up with that inequality? · 4 years, 1 month ago