# Pre-Cal.

posted by .

f(x)=x^3-3x^2+x-12
=x^2(x-3)+4(x-3)
=x^2+4) (x-3)
so if f(x) = 0 at x = 3,+/- 2i.
what is the f(x) factored form

• Pre-Cal. -

(x)=x^3-3x^2+x-12
=x^2(x-3)+4(x-3) Not true

That would be
x^3 - 3x^2 + 4x - 12
=========================
You have told me that the zeros of the function are:
3
2i
-2i
Therefore the factored form is:
(x-3)(x-2i)(x+2i)
try multiplying that out

• Pre-Cal. -

that is
(x^2+4)(x-3)
which is
x^3 - 3 x^2 + 4x -12
so I think you have a typo in your opening line

• Pre-Cal. -

where is the typo in which opening line? And I believe I used the wrong symbol for the factored function it's suppose to be a plus sign over a __ but I can't figure how to type this. and maybe that's what is giving us the answer you have given me. Could that be it? and that is not -12, negative 12 it's minus 12

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