x^n+a 1*x^(n- 1)+.+a(n- 1)*x+an=0。 ( 1)
In other words, the coefficient of the first term is 1. If it is not 1, both sides are divided by the first coefficient.
Let its root be x 1, x2,., xn.
Then the above equation can be expressed as
(x-x 1)*(x-x2)。 *(x-xn)=0。 (2)
Expand formula (2) and compare the coefficient with formula (1), and you can get
x 1+x2+。 xn=-a 1
x 1*x2+x 1*x3+...+x(n- 1)*xn=a2
x 1*x2*x3+x 1*x2*x4+...=-a3
x 1*x2*。 xn=(- 1)^n*an