高二数学， x³-12x+16=0这样的怎么求啊
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x^3-12x+16=0
f(x) =x^3-12x+16
f(2) = 0
x-2 is a factor
f(x) = (x-2)(x^2+ax+b)
coef. of constant
-2b = 16
b=-8
put x=1
5 = -(-7+a)
a=2
f(x) =(x-2)(x^2+2x-8)
=(x-2)(x+4)(x-2)
x^3-12x+16=0
(x-2)(x+4)(x-2)=0
x=2 or -4
f(x) =x^3-12x+16
f(2) = 0
x-2 is a factor
f(x) = (x-2)(x^2+ax+b)
coef. of constant
-2b = 16
b=-8
put x=1
5 = -(-7+a)
a=2
f(x) =(x-2)(x^2+2x-8)
=(x-2)(x+4)(x-2)
x^3-12x+16=0
(x-2)(x+4)(x-2)=0
x=2 or -4
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