Posted by allison--Please help!! on Thursday, December 27, 2012 at 5:08pm.
You factored the cubic polynomial correctly as
(x-3)(x^2 + 16)
The first term tells the function has a zero at x = 3.
The other roots are imaginary. x = +/- 4i
yes, I did put:
x^2+16=0
x^2= -16
x= + or -4i
But she wrote that on the first part that I submitted and I don't really know what it is that she wants me to do.
**So are you saying that the value for x is 3, but then what is the zero for the polynomial based on the first factor??
x-3 = 0
means x = 0
so the polynomial is zero at x = 3
also at - 4 i and at + 4 i
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