# How do you write the polynomial function given the following zeros: -5i, -i, i, 2i?

It is $P \left(x\right) = \left(x + 5 i\right) \left(x + i\right) \left(x - i\right) \left(x - 2 i\right) = {x}^{4} + 3 i {x}^{3} + 11 {x}^{2} + 3 i x + 10$

where $i$ is the imaginery unit (${i}^{2} = - 1$)

Sep 17, 2015

If the polynomial also has Real coefficients, then it also has zeroes $5 i$ and $- 2 i$

The simplest such polynomial is:

$f \left(x\right) = \left(x - 5 i\right) \left(x + 5 i\right) \left(x - i\right) \left(x + i\right) \left(x - 2 i\right) \left(x + 2 i\right)$

$= {x}^{6} + 30 {x}^{4} + 129 {x}^{2} + 100$

#### Explanation:

The Complex zeroes of polynomials with Real coefficients always occur in conjugate pairs.

So in our example, since $- 5 i$ and $2 i$ are zeroes, so are $5 i$ and $- 2 i$.

Hence:

$f \left(x\right) = \left(x - 5 i\right) \left(x + 5 i\right) \left(x - i\right) \left(x + i\right) \left(x - 2 i\right) \left(x + 2 i\right)$

$= \left({x}^{2} + 25\right) \left({x}^{2} + 1\right) \left({x}^{2} + 4\right)$

$= {x}^{6} + \left(25 + 1 + 4\right) {x}^{4} + \left(1 \cdot 4 + 25 \cdot 4 + 25 \cdot 1\right) {x}^{2} + \left(25 \cdot 1 \cdot 4\right)$

$= {x}^{6} + 30 {x}^{4} + 129 {x}^{2} + 100$

Any polynomial in $x$ with these roots will be a multiple (scalar or polynomial) of this $f \left(x\right)$.