# How do you write a polynomial function given the real zeroes -2,-2,3,-4i and coefficient 1?

Jan 5, 2016

${x}^{5} + {x}^{4} + 8 {x}^{3} + 4 {x}^{2} - 128 x - 192$

#### Explanation:

The major trick with this problem is remembering that complex roots always come in pairs.

Thus, along with the root of $- 4 i$, the polynomial will also have a root of $4 i$.

The polynomial can be written as:

${\left(x + 2\right)}^{2} \left(x - 3\right) \left(x + 4 i\right) \left(x - 4 i\right)$

$= \left({x}^{2} + 4 x + 4\right) \left(x - 3\right) \left({x}^{2} + 16\right)$

When distributed completely, this gives

${x}^{5} + {x}^{4} + 8 {x}^{3} + 4 {x}^{2} - 128 x - 192$

graph{x^5+x^4+8 x^3+4 x^2-128 x-192 [-10, 10, -500, 301.6]}

As you can see, the graph has an odd degree $\left(5\right)$, has a root of $- 2$ with multiplicity $2$ and a root at $3$.