Question

x的三次方-7x的平方+4x+2用待定系数法怎么解求过程

Answer 1

f(x)=x^3-7x^2+4x+2=0 (1)

(x-a)(x^2+bx+c)=0

x^3+bx^2+cx-ax^2-abx-ac=0

x^3+(b-a)x^2+(c-ab)x-ac=0

b-a=-7

c-ab=4

ac=-2

从(1)看出：x=1为(1)的实根；即：a=1 b=-6 c=-2

(1)的因式分f(x)=(x-1)(x^2-6x-2)=(x-1)(x-3-√11)(x-3+√11)

(1)的根：x1=1,x2=3+√11 ,x3=3-√11