Write the general monic degree-n equation as
xn+p1xn−1+p2xn−2+⋯+pn−1x+pn=0,
and let α1,α2,…,αn be its roots. Comparing this with the factored form (x−α1)(x−α2)⋯(x−αn) and expanding, the coefficient of each power of x turns out to be (up to a sign) one of the elementary combinations of the roots. Writing sk for the sum of all products of the roots taken k at a time, the pattern is
s1=∑αi=−p1,s2=∑i<jαiαj=p2,s3=∑i<j<kαiαjαk=−p3, …,sn=α1α2⋯αn=(−1)npn.
In words: the signs of s1,s2,s3,… alternate starting from a minus sign, and each sk equals ±pk. For a cubic x3+p1x2+p2x+p3=0 with roots α,β,γ this reads α+β+γ=−p1, αβ+βγ+γα=p2, αβγ=−p3 — the same relations you already used for quadratics, just extended one degree further.
These relations are powerful in both directions. Given an equation, you can read off sums and products of the roots without solving it. Given the roots (or some of them), you can reconstruct the equation, because xn+p1xn−1+⋯+pn≡(x−α1)⋯(x−αn) as polynomials, so simply multiplying out the factors gives the coefficients directly.
Worked example. If 2,−1,3,−2 are the roots of x4+ax3+bx2+cx+d=0, find a,b,c,d. …