# How do you write a polynomial function of least degree that has real coefficients, the following given zeros 8,-i, i and a leading coefficient of 1?

Nov 12, 2016

${x}^{4} - 9 {x}^{3} + 9 {x}^{2} - 9 x + 8$

#### Explanation:

As the two zeros among four zeros, viz. $8$, $- i$, $i$ and $1$ have two complex numbers, which are complex conjugate of each other, it is possible to write a polynomial function of least degree (which is $4$ as there are $4$ zeros with no multiplicity, i.e. repeat zeros) that has real coefficients.

A function $f \left(x\right)$ with zeros $a$, $b$, $c$ and $d$ is $\left(x - a\right) \left(x - b\right) \left(x - c\right) \left(x - d\right)$ and hence the function with $8$, $- i$, $i$ and $1$as zeros, will be

$\left(x - 8\right) \left(x - \left(- i\right)\right) \left(x + i\right) \left(x - 1\right)$

= $\left(x - 8\right) \left(x + i\right) \left(x - i\right) \left(x - 1\right)$

= $\left(x - 8\right) \left(x - 1\right) \left({x}^{2} + i x - i x + \left(- {i}^{2}\right)\right)$

= $\left(x - 8\right) \left(x - 1\right) \left({x}^{2} + 1\right)$ as ${i}^{2} = - 1$

= $\left({x}^{2} - 9 x + 8\right) \left({x}^{2} + 1\right)$

= ${x}^{4} - 9 {x}^{3} + 8 {x}^{2} + {x}^{2} - 9 x + 8$

= ${x}^{4} - 9 {x}^{3} + 9 {x}^{2} - 9 x + 8$