# How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are 4, 4, 2+i?

Sep 17, 2017

$f \left(x\right) = {x}^{4} - 12 {x}^{3} + 53 {x}^{2} - 104 x + 80$

#### Explanation:

Assuming you want a polynomial with real coefficients, the complex conjugate $2 - i$ must also be a zero and the monic polynomial of least degree is:

$f \left(x\right) = \left(x - 4\right) \left(x - 4\right) \left(x - \left(2 + i\right)\right) \left(x - \left(2 - i\right)\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left({x}^{2} - 8 x + 16\right) \left(\left(x - 2\right) + i\right) \left(\left(x - 2\right) - i\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left({x}^{2} - 8 x + 16\right) \left({\left(x - 2\right)}^{2} - {i}^{2}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left({x}^{2} - 8 x + 16\right) \left({x}^{2} - 4 x + 4 + 1\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left({x}^{2} - 8 x + 16\right) \left({x}^{2} - 4 x + 5\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{4} - 12 {x}^{3} + 53 {x}^{2} - 104 x + 80$

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