Find all zeros of the following polynomial. Write the polynomial in factored form. f(x)=x^3-3x^2+16x-48

Question

I put: x^2(x-3)+16(x=3)
(x-3)(x^2+16)
For zeros:
x-3=0
x=0
**My teacher stated check the equation solution again. What is the value for x and hence what is the zero for the polynomial based on the first factor?

**Can someone help me?

Answer 1

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

Answer 2

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??

Answer 3

x-3 = 0
means x = 0
so the polynomial is zero at x = 3
also at - 4 i and at + 4 i