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PMO National Stage Oral Phase

Algebra Difficulty 3.8 AMC 10/12 Find the answer Philippines

Problem:
Let f(x)f(x) be a polynomial function of degree 20162016 whose 20162016 zeroes have a sum of SS. Find the sum of the 20162016 zeroes of f(2x−3)f(2x-3) in terms of SS.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
Let r1,r2,⋯ ,r2016r_{1}, r_{2}, \cdots, r_{2016} be the zeroes of f(x)f(x). We can then write ff as
f(x)=c(x−r1)(x−r2)⋯(x−r2016) f(x) = c\left(x - r_{1}\right)\left(x - r_{2}\right) \cdots \left(x - r_{2016}\right)
where ∑i=12016ri=S\sum_{i=1}^{2016} r_{i} = S. Thus
f(2x−3)=c(2x−3−r1)(2x−3−r2)⋯(2x−3−r2016) f(2x-3) = c\left(2x-3 - r_{1}\right)\left(2x-3 - r_{2}\right) \cdots \left(2x-3 - r_{2016}\right)
which has zeroes
r1+32,r2+32,…,r2016+32 \frac{r_{1} + 3}{2}, \frac{r_{2} + 3}{2}, \ldots, \frac{r_{2016} + 3}{2}
This means that the required sum is
∑i=12016ri+32=12[∑i=12016ri+3×2016]=12S+3024 \sum_{i=1}^{2016} \frac{r_{i} + 3}{2} = \frac{1}{2}\left[\sum_{i=1}^{2016} r_{i} + 3 \times 2016\right] = \frac{1}{2} S + 3024

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.