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Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 18G / a / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 4 18G a
Created 2026-09-24 Updated 2026-09-29  0 By others on same topic  0 Discussions Create my own version
Let f(x)=a∏i=1n​(x−αi​). Its polynomial discriminant is
Δ(f)=a2n−2∏i<j​(αi​−αj​)2=(−1)n(n−1)/2an−2Res(f,f′).
(1)
The squared Vandermonde determinant is a symmetric polynomial in the roots. By the Fundamental theorem of symmetric polynomials, it is a polynomial in their elementary symmetric functions, which are the coefficients of f by Vieta formulas; hence Δ(f)∈K. Equivalently, the resultant formula puts it directly in K.
For f=x4+rx+s, a resultant calculation with f′=4x3+r gives
Δ(f)=256s3−27r4​.
(2)

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