Question

How do you find all the zeros of `p(x)=(x^2+9)(x+3)`?

Answer 1

`x=-3`

Explanation:

`x^2+9=0, x+3=0`
`x^2=-9, x=-3`
`x=+-sqrt-9-> no solution, x=-3`
`x=-3`
If it is in the set of complex numbers then answers will be
`x=+-3i or x=-3`