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
(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
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
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
To find the factored form of f(x), we need to determine the factors of the expression.
The original expression is:
f(x) = x^3 - 3x^2 + x - 12
By factoring out a common factor, x - 3, we can rewrite the expression as:
f(x) = (x - 3)(x^2 + 4)
So, the factored form of f(x) is (x - 3)(x^2 + 4).