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Let f(x)=(x^2+6x+9)^50-4x+3 and let r_1,r_2,r_3...r_100 be the roots of f(x).

Compute (r_1+3)^100 + (r_2+3)^100 + ... + (r_100+3)^100

 

Could you please explain? 

Thank you

 Jun 18, 2020
edited by Guest  Jun 18, 2020
 #1
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 The answer is -1500

 Jun 18, 2020
 #2
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How do I solve it?

 Jun 18, 2020
 #3
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+1

\(f(x) = (x^2 + 6x + 9)^{50} - 4x + 3 = (x + 3)^{100} - 4x + 3\)

 

Substituting \(r_1, r_2,\cdots,r_{100}\) into f(x) = 0, we have

 

\(\begin{cases}\left(r_1 + 3\right)^{100} - 4r_1 + 3 = 0\\\left(r_2 + 3\right)^{100} - 4r_2 + 3 = 0\\\cdots\\\left(r_{100} + 3\right)^{100} - 4r_{100} + 3 = 0\end{cases}\)

 

Moving all the terms to the right hand side and adding them up,

 

\((r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100} = 4(r_1 + r_2 + \cdots + r_{100}) - 300\)

 

That means the required answer is 4(sum of roots) - 300.

 

By Vieta's formula, \(\text{sum of roots} = -\dfrac{\text{coefficient of }x^{99}}{\text{coefficient of }x^{100}}\).

 

Using binomial theorem, the coefficient of x100 is 1 and the coefficient of x99 is 300.

 

Therefore \(r_1 + r_2 + \cdots + r_{100} = -300\).

 

Therefore \((r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100} = 4(-300) - 300 = \boxed{-1500}\)

 Jun 18, 2020

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