# How do you write a polynomial with zeros 6, 2 + 2i?

Mar 1, 2016

$p \left(x\right) = {x}^{3} - 10 {x}^{2} + 32 x - 48$

#### Explanation:

If a polynomial has a zero at $a$, then it also has a factor of $\left(x - a\right)$.

Here, we see that the zero at $6$ corresponds to a factor of $\left(x - 6\right)$, but we need to account for the complex zero.

In polynomials with real coefficients, which I'm assuming this is, complex roots always come in pairs. The pairs of complex roots are always complex conjugates of one another. This means, that if $2 + 2 i$ is a zero, $2 - 2 i$ will be as well.

Constructing a polynomial with zeros of $6 , 2 + 2 i , 2 - 2 i$, we obtain:

$p \left(x\right) = \left(x - 6\right) \left(x - \left(2 + 2 i\right)\right) \left(x - \left(2 - 2 i\right)\right)$

Multiplying the $\left(x - \left(2 + 2 i\right)\right) \left(x - \left(2 - 2 i\right)\right)$ is odd, but we can manipulate it to be more friendly:

$p \left(x\right) = \left(x - 6\right) \left(\left(x - 2\right) - 2 i\right) \left(\left(x - 2\right) + 2 i\right)$

Now, the final two terms are in the form $\left(a - b\right) \left(a + b\right)$, which equals ${a}^{2} - {b}^{2}$.

$p \left(x\right) = \left(x - 6\right) \left({\left(x - 2\right)}^{2} - {\left(2 i\right)}^{2}\right)$

$p \left(x\right) = \left(x - 6\right) \left({x}^{2} - 4 x + 4 - 4 {i}^{2}\right)$

Recall that ${i}^{2} = - 1$.

$p \left(x\right) = \left(x - 6\right) \left({x}^{2} - 4 x + 4 + 4\right)$

$p \left(x\right) = \left(x - 6\right) \left({x}^{2} - 4 x + 8\right)$

Distribute completely:

$p \left(x\right) = {x}^{3} - 10 {x}^{2} + 32 x - 48$