# How do you write a polynomial in standard form given zeros 1 - 2i and 1 + 2i?

Apr 2, 2016

$f \left(x\right) = {x}^{2} - 2 x + 5$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Use this with $a = \left(x - 1\right)$ and $b = 2 i$ as follows:

$f \left(x\right) = \left(x - 1 - 2 i\right) \left(x - 1 + 2 i\right)$

$= \left(\left(x - 1\right) - 2 i\right) \left(\left(x - 1\right) + 2 i\right)$

$= {\left(x - 1\right)}^{2} - {\left(2 i\right)}^{2}$

$= {x}^{2} - 2 x + 1 + 4$

$= {x}^{2} - 2 x + 5$

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