How do you find the zeros of the function $F (x) = x^{3} - 3 x^{2} + 4 x - 2$ $F(x)=x^3-3x^2+4x-2$ ?

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Answer 1 · 2015-05-12T00:59:02

Reminder:

When a + b + c + d = 0, then the function has (x - 1) as a factor.

$F (x) = (x - 1) (x^{2} - 2 x + 2)$ $F(x) = (x - 1)(x^2 - 2x + 2)$

Since the trinomial in parentheses can't be factored (D = -4 < 0),

then there is one unique zero: x = 1.

How do you find the zeros of the function F(x)=x3−3x2+4x−2F(x)=x^3-3x^2+4x-2?