Question

Use the given zero to find all the zeros of the function, Please explain I do not understand?

Given zero: 3i

(The i is imaginary)

Function: f(x)=x³+x²+9x+9

Answer 1

`{-3i, 3i, -1}`

Explanation:

The zeros, or roots, of a function `f(x)` are the solutions to the equation `f(x) = 0`. I.e. they are the values which return `0`, when entered into `f(x)`.

This problem uses two properties of roots.

• `x_0` is a zero of a polynomial `P(x)` if and only if `(x-x_0)` is a factor of `P(x)`.

• If `P(x)` is a polynomial with real coefficients and `z=a+bi` is a nonreal complex number which is a zero of `P(x)`, then its complex conjugate `bar(z)=a-bi` is also a zero of `P(x)`.

As `f(x) = x^3+x^2+9x+9` is a polynomial with real coefficients, and `3i = 0+3i` is a zero of `f(x)`, then the second property gives us that `0-3i=-3i` must also be a zero of `f(x)`.

Because `3i` and `-3i` are zeros of `f(x)`, the first property gives us that `(x+3i)` and `(x-3i)` are both factors of `f(x)`. Thus their product `(x+3i)(x-3i) = x^2+9` is also a factor of `f(x)`.

As `f(x)` is a degree `3` polynomial, it must have exactly `3` zeros (with the possibility of some repeating). We have two of them, so let's let `x_0` be the third zero. Then we know `(x-x_0)` is a factor of `f(x)`, and it is indeed the last nonconstant factor, so for some nonzero `c`,

`x^3+x^2+9x+9= c(x+3i)(x-3i)(x-x_0)`

`=c(x^2+9)(x-x_0)`

`=cx^3-cx_0x^2+9cx-9cx_0`

Finally, we equate coefficients to find the unknown values. Equating the coefficients of the `x^3` terms, we have `1=c`. Equating the coefficients of the constant terms, we have `9=-9cx_0 = -9x_0`. Thus `x_0 = -1`

So, all together, `f(x)` has the zeros `{-3i, 3i, -1}`