# How do you write a polynomial with zeros 2, -2, -6i and leading coefficient 1?

May 10, 2017

Identify the four (not three) zeroes and write out the zeroes in factored form. Answer: ${x}^{4} + 32 {x}^{2} - 144 = 0$

#### Explanation:

Original question: Find polynomial with zeroes $2 , - 2 , - 6 i$ and leading coefficient of $1$

We know that each complex zero must have a pair that is also a complex zero, which is its conjugate. Therefore, if $- 6 i$ is a zero, then $6 i$ is a zero too.

We also know that zeroes of a polynomial come in the form $\left(x - a\right) \left(x - b\right) \left(x - c\right) \left(x - d\right) = 0$ where $a , b , c , d$ are zeroes, therefore we can write our polynomial in this same form:
$\left(x - 2\right) \left(x + 2\right) \left(x + 6 i\right) \left(x - 6 i\right) = 0$

Now, we can write out the expanded form by using the fact that $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ and expand the resulting product of binomials:
Note that ${\left(6 i\right)}^{2} = 36 \left(- 1\right) = - 36$
$\left({x}^{2} - 4\right) \left({x}^{2} + 36\right) = 0$
${x}^{2} \cdot {x}^{2} + {x}^{2} \cdot 36 - 4 \cdot {x}^{2} - 4 \cdot 36 = 0$
${x}^{4} + 36 {x}^{2} - 4 {x}^{2} - 144 = 0$

Combining like-terms, we get:
${x}^{4} + 32 {x}^{2} - 144 = 0$