Question

一元四次方程

p(x)=x^4-6x^2+3x+2
求零点...
需要详细的解题步骤啊..

最好写上解这些题的习惯做法,简便方法..
答案好的,加20分...

谢谢..
一楼的,不是很懂你的做法,为什么一开始有p(1)呢,是固定解法还是怎样,能不能具体的解释解释你的解法..
还有这一类题目的习惯做法是什么 啊..?
谢谢了..

Answer 1

p(x)=x^4-6x^2+3x+2
p(1)=1-6+3+2 = 0
x= 1 is root
let p(x) = (x-1)(x^3+bx^2+cx+d)

compare the coef. of constant
-d = 2 => d =-2
compare the coef. of x^3
-1+b = 0
b = 1
compare the coef of x
-c+ d = 3
-c-2 =3
c = -5

p(x)=(x-1)(x^3+x^2-5x-2)
p(2)=0
x=2 is root

let
(x^3+x^2-5x-2) = (x-2)(x^2+ex+f)
compare the coef. of constant
-2f=-2
f=1
compare the coef of x^2
e-2 =1
e = 3

(x^3+x^2-5x-2) = (x-2)(x^2+3x+1)

p(x)=(x-1)(x-2)(x^2+3x+1)

x^2+3x+1 = 0
x = (-3+√5) / 2 or (-3-√5) / 2

零点: 1,2,(-3+√5) / 2 , (-3-√5) / 2

附加：p(1）=0，这不是固定的解法，这只适合这个方程而已。