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In the AOPS vol 2 problem solving book, it states that you can find the sum of the reciprocals of a polynomial by flipping the coefficients(first -> last, last -> first etc). The book summarized the fact that you can "transform polynomials" to get a desired sum of roots (just like the example I mentioned above). However, I'm confused on how to do it for more complicated sums, so could anyone explain the general method of transforming polynomials.

Here is an example question to demonstrate:

If the roots of polynomial $x^4−3x^3−27x^2−13x+42$ are $r_1$, $r_2$, $r_3$, $r_4$, find $(1/(r_1 + 1))$+$(1/(r_2 + 1))$+$(1/(r_3 + 1))$+$(1/(r_4 + 1))$.

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A general method does not exist, I think. You can find some useful rules here https://en.wikipedia.org/wiki/Vieta%27s_formulas . I will help you solving your particular example, step by step.

Call $f(x)= x^4-3x^3-27x^2-13x+42$, and let $r_1, \dots, r_4$ be the roots of $f$.

Then $r_1 +1 = s_1, \dots, r_4+1 = s_4$ are the roots of the polynomial $f(x-1)$ since $f(s_i-1)= f(r_i) 0$. Let's compute $$f(x-1) = (x-1)^4-3(x-1)^3-27(x-1)^2-13(x-1)+42 =$$ $$= \mbox{ tedious computations } = x^4-7x^3 - 12x^2+28x+32$$

Now, $\frac{1}{s_1} = z_1 , \dots, \frac{1}{s_4} = z_4$ are the roots of $$32x^4+28x^3-12x^2-7x+1$$ simply swapping coefficients.

Finally, you should know (this is one of Vieta's formulas) that the sum of roots of a polynomial $\sum_{0 \le k \le n} a_kx^k$ is $-\frac{a_{n-1}}{a_n}$, so the final result is $$\sum_i z_i = -\frac{28}{32} = -\frac{7}{8}$$

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  • $\begingroup$ Excellent walkthrough! :D $\endgroup$ – Sujoy Purkayastha Aug 29 '15 at 0:08

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