Question

求所有以有理数a、b、c为根的三次多项式x^3+ax^2+bx+c.

Answer 1

利用根与系数的关系.

a+b+c=-a =>c=-2a-b (1)

ab+bc+ca=b =>ab+(a+b)(-2a-b)=b =>-2a^2-3ab-b^2=b (2)

abc=-c =>ab=-1 或者c=0 (3)

(3)代入(2)

c=0 =>ab=b =>a=1或者b=0

a=1代入(1) =>b=-2

b=0 =>a=0

ab=-1=>

2/b^2+3-b^2=b

=>2+3b^2-b^3-b^4=0

没有有理数根.

所以：a=0 b=0 c=0

a=1 b=-2 c=0