h

Question

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anonymous · Answer

First, can you factor the equation at all?

anonymous · Answer

$$ f_{x} = x^4 - 4x^3+3x^2$$

What's similar about each term?

anonymous · Answer

Correct!  But each of them have a power of at least $$x^2$$

so you can pull them out front
$$f_{x} = x^2(x^2 - 4x+3)$$

anonymous · Answer

Yeah, exactly.  So you'll have 4 solutions for x.
x = 0
x = 0
from the x^2

Then two solutions from factoring the polynomial in the parenthesis

anonymous · Answer

Yup.  Either b or d

anonymous · Answer

Then
$$x^2 - 4x + 3$$
can be factored into 

$$(x-3)(x-1)$$

anonymous · Answer

set those both equal to zero and solve for x

anonymous · Answer

nope.  The zeroes are for when you solve the equation.  as in "If x were this number, this term would be zero"

if x = -3, then in your expression

(-3 -3) = -6 ≠ 0

So for

(x-3) = 0
x = 3

anonymous · Answer

So B

anonymous · Answer

sorry for the tardiness :/