Let $f ( x ) = x ^ { 3 } - 3 x ^ { 2 } + 2 x + 2$ For what value(s) of x is $f ( x ) = 2$ ?
A) $x = 1 , x = 3$
B) $x = 4 , x = 1 , x = 5$
C) $x = 2 , x = 1 , x = 1$
D) $x = 0 , x = 2$
E) $x = 0 , x = 1 , x = 2$

Question

Quizplus · Accepted Answer

By substituting $f(x) = 2$, we get the equation:
$$x^3-3x^2+2x=0$$
We can factor an x out of the equation to get:
$$x(x^2-3x+2)=0$$
Solving for x, we get x = 0, x = 1, or x = 2. Therefore, the answer is choice E.

Let f(x)=x3−3x2+2x+2f ( x ) = x ^ { 3 } - 3 x ^ { 2 } + 2 x + 2f(x)=x3−3x2+2x+2

Let $f ( x ) = x ^ { 3 } - 3 x ^ { 2 } + 2 x + 2$