Roots of a 100-th degree polynomial

Algebra Level 5

f ( x ) = x 100 + a 99 x 99 + … + a 1 x + a 0 \large f(x)=x^{100}+a_{99}x^{99}+\ldots+a_{1}x+a_{0}

Let f ( x ) f(x) be a 100th degree polynomial such that a 99 = − 10 , a_{99}=-10, a 98 = 5 a_{98}=5 and a 97 = − 2 a_{97}=-2 with roots x 1 , x 2 , x 3 , … x 100 x_{1}, x_{2}, x_{3}, \ldots x_{100} . Evaluate ∑ k = 1 100 x k 3 \displaystyle \sum_{k=1}^{100} x_{k}^3 .


The answer is 856.

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Arturo Presa by Arturo Presa , 62, USA

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3 solutions

Arturo Presa
Aug 8, 2015

Using Vieta's formulas we have that − a 99 = ∑ k = 1 100 x k = 10 , -a_{99}=\sum_{k=1}^{100}{x_{k}}=10, a 98 = ∑ 1 ≤ k < l ≤ 100 x k x l = 5 , a_{98}=\sum_ {1\leq k<l\leq 100}{x_{k}x_{l}}=5, and − a 97 = ∑ 1 ≤ k < l < m ≤ 100 x k x l x m = 2. -a_{97}=\sum_{1\leq k<l<m\leq 100}{x_{k}x_{l}x_{m}}=2.

Then ∑ k = 1 100 x k 2 = ( ∑ k = 1 100 x k ) 2 − 2 ∗ ∑ 1 ≤ k < l ≤ 100 x k x l = 1 0 2 − 2 ∗ 5 = 90. \sum_{k=1}^{100} x_{k}^2=(\sum_{k=1}^{100}x_{k})^2-2 *\sum_ {1\leq k<l\leq 100}{x_{k}x_{l}}=10^{2}-2*5=90. Therefore ∑ k = 1 100 x k 3 = ∑ k = 1 100 x k 3 − 3 ∗ ∑ 1 ≤ k < l < m ≤ 100 x k x l x m + 6 = \sum_{k=1}^{100} x_{k}^3= \sum_{k=1}^{100} x_{k}^3 -3 *\sum_ {1\leq k<l<m\leq 100}{x_{k}x_{l}x_{m}} +6= = ( ∑ k = 1 100 x k ) ( ∑ k = 1 100 x k 2 − ∑ 1 ≤ k < l ≤ 100 x k x l ) + 6 = 10 ∗ ( 90 − 5 ) + 6 = 856 =(\sum_{k=1}^{100}x_{k})(\sum_{k=1}^{100} x_{k}^2-\sum_ {1\leq k<l\leq 100}{x_{k}x_{l}})+6=10*(90-5)+6=856

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Love this problem and the solution. :)

Revanth Gumpu - 5 years, 10 months ago

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Thank you @RevanthGumpu !

Arturo Presa - 5 years, 10 months ago
Chew-Seong Cheong
Aug 17, 2015

Using Vieta's formulas, we have:

{ P n = ∑ k = 1 100 x k n S 1 = ∑ k = 1 100 x k = − a 99 = 10 S 2 = ∑ c y c l x j x k = a 98 = 5 S 3 = ∑ c y c l x i x j x k = − a 97 = 2 \begin{cases} P_n = \displaystyle \sum_{k=1}^{100} x_k^n \\ S_1 = \displaystyle \sum_{k=1}^{100} x_k = - a_{99} = 10 \\ S_2 = \displaystyle \sum_{cycl} x_jx_k = a_{98} = 5 \\ S_3 = \displaystyle \sum_{cycl} x_ix_jx_k = -a_{97} = 2 \end{cases}

P 1 = S 1 = 10 P 2 = S 1 P 1 − 2 S 2 = 10 ( 10 ) − 2 ( 5 ) = 90 P 3 = S 1 P 2 − S 2 P 1 + 3 S 3 ⇒ ∑ k = 1 100 x k 3 = 10 ( 90 ) − 5 ( 10 ) + 3 ( 2 ) = 856 \begin{aligned} P_1 & = S_1 = 10 \\ P_2 & = S_1 P_1 - 2S_2 = 10(10) - 2(5) = 90 \\ P_3 & = S_1 P_2 - S_2 P_1 + 3S_3 \\ \Rightarrow \sum_{k=1}^{100} x_k^3 & = 10(90) - 5(10) + 3(2) = \boxed{856} \end{aligned}

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i did same...

Dev Sharma - 5 years, 9 months ago
Akshat Sharda
Sep 15, 2015

Newton's Sum is great!!

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