30x^4+60x^3-1110x^2+300x+5040=0的解

Question

这个四次方程求大神解一解

Answer 1

30x^4+60x^3-1110x^2+300x+5040=0
x^4+2x^3-37x^2+10x+168=0

f(x) =x^4+2x^3-37x^2+10x+168
f(3) =0
x^4+2x^3-37x^2+10x+168 = (x-3)(x^3+ax^2+bx-56)

coef. of x
-56-3b=10
b=-22

coef. of x^2
b-3a =-37
a =5

x^4+2x^3-37x^2+10x+168 = (x-3)(x^3+5x^2-22x-56)

g(x) =x^3+5x^2-22x-56
g(4)= 0

x^3+5x^2-22x-56 = (x-4)(x^2+cx -14)
coef. of x
-14-4c=-22
c=2

x^3+5x^2-22x-56
= (x-4)(x^2+2x -14)

30x^4+60x^3-1110x^2+300x+5040=0
x^4+2x^3-37x^2+10x+168=0
(x-3)(x-4)(x^2+2x -14) =0
x=3 or 4 or -1+√15 or -1-√15

Answer 2

化简到x^4+2x^3-37x^2+10x+168=0
分解(x-3)(x-4)(x+2)(x+7)=0
得:3,4,-2,-7追问

汗...代进去算了一下你是对的...手快给了上面那个人不好意思

Answer 3

给你截图了请采纳追答