# How do you find a polynomial f(x) of degree 3 with real coefficients and the following zeros 1, 3-i?

Nov 18, 2015

Since complex roots ALWAYS occur in conjugate pairs ...

#### Explanation:

$f \left(x\right) = \left(x - 1\right) \left[x - \left(3 - i\right)\right] \left[x - \left(3 + i\right)\right]$

Next, simplify ...

$f \left(x\right) = \left(x - 1\right) \left({x}^{2} - 6 x + 10\right)$

$= {x}^{3} - 7 {x}^{2} + 16 x - 10$

From the graph below, the function is a cubic with only one real root (at $x = 1$) and two imaginary roots .

hope that helped