# 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

Dec 4, 2016

$\left\{- 3 i , 3 i , - 1\right\}$

#### Explanation:

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

This problem uses two properties of roots.

• ${x}_{0}$ is a zero of a polynomial $P \left(x\right)$ if and only if $\left(x - {x}_{0}\right)$ is a factor of $P \left(x\right)$.

• If $P \left(x\right)$ is a polynomial with real coefficients and $z = a + b i$ is a nonreal complex number which is a zero of $P \left(x\right)$, then its complex conjugate $\overline{z} = a - b i$ is also a zero of $P \left(x\right)$.

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

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

As $f \left(x\right)$ 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 $\left(x - {x}_{0}\right)$ is a factor of $f \left(x\right)$, and it is indeed the last nonconstant factor, so for some nonzero $c$,

${x}^{3} + {x}^{2} + 9 x + 9 = c \left(x + 3 i\right) \left(x - 3 i\right) \left(x - {x}_{0}\right)$

$= c \left({x}^{2} + 9\right) \left(x - {x}_{0}\right)$

$= c {x}^{3} - c {x}_{0} {x}^{2} + 9 c x - 9 c {x}_{0}$

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 $9 = - 9 c {x}_{0} = - 9 {x}_{0}$. Thus ${x}_{0} = - 1$

So, all together, $f \left(x\right)$ has the zeros $\left\{- 3 i , 3 i , - 1\right\}$