Polynomial problem

Let g(x)g(x) be a polynomial such that g(x)=x84x7+7x6+ax5+bx4+cx3+dx2+ex+fg(x)=x^{8}-4x^{7}+7x^{6}+ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f factorizes into 88 linear factors.Also roots of the polynomial are all positive,where a,b,c,d,e,fa,b,c,d,e,f are all real numbers.Sum of all possible values of ff can be written as ab\frac{a}{b},where aa and bb are co-prime positive integers.Find a+b?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
6 years, 2 months ago

No vote yet
2 votes

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Perhaps they rejected it because you use a,ba,b for the coefficients of g(x)g(x) as well as the value of ff.

By Vieta's Relations, the sum of the 88 roots is 44 and the sum of the squares is 4227=24^2 - 2\cdot 7 = 2.

Hence, the arithmetic mean of the roots is 48=12\dfrac{4}{8} = \dfrac{1}{2} and the root-mean-square of the roots is 28=12\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 12\dfrac{1}{2}.

Thus, g(x)=(x12)8g(x) = \left(x - \tfrac{1}{2}\right)^8 is the unique polynomial g(x)g(x). So, f=g(0)=128=1256a+b=257f = g(0) = \dfrac{1}{2^8} = \dfrac{1}{256} \leadsto a+b = \boxed{257}.

Jimmy Kariznov - 6 years, 2 months ago

Log in to reply

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

Kishan k - 6 years, 2 months ago

Log in to reply

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

Thanks!

Pranav Arora - 6 years, 2 months ago

Log in to reply

The RMS-AM Inequality states that if x1,,xn0x_1, \ldots, x_n \ge 0, then 1nk=1nxk21nk=1nxk\sqrt{\dfrac{1}{n}\displaystyle\sum_{k = 1}^{n}x^2_k} \ge \dfrac{1}{n}\displaystyle\sum_{k=1}^{n}x_k.

Furthermore, 1nk=1nxk2=1nk=1nxk=a\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 x1==xn=ax_1 = \cdots = x_n = a.

In this problem, that theorem allows us to conclude that the roots of g(x)g(x) are all equal to 12\tfrac{1}{2}.

This shows that there is only one possible polynomial g(x)g(x), namely g(x)=(x12)8g(x) = \left(x - \tfrac{1}{2}\right)^8.

Jimmy Kariznov - 6 years, 2 months ago

Log in to reply

@Jimmy Kariznov Awesome as always, thanks Jimmy! :)

Mind throwing some light on how you instantly came up with that inequality?

Pranav Arora - 6 years, 2 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...