# How do you find a third degree polynomial given roots 5 and 2i?

Nov 7, 2016

Please see the explanation section below.

#### Explanation:

For a third degree polynomial, we need 3 linear factors.

Since $5$ and $2 i$ are roots (zeros), we know that $x - 5$ and $x - 2 i$ are factors.

If we want a polynomial with real coeficients, then the complex conjugate of $2 i$ (which is $- 2 i$) must also be a root and $x + 2 i$ must be a factor.

One polynomial with real coefficients that meets the requirements is

$\left(x - 5\right) \left(x - 2 i\right) \left(x + 2 i\right) = \left(x - 5\right) \left({x}^{2} + 4\right)$

$= {x}^{3} - 5 {x}^{2} + 4 x - 20$

Any constant multiple of this also meets the requirements.

For example

$7 \left({x}^{3} - 5 {x}^{2} + 4 x - 20\right) = 7 {x}^{3} - 35 {x}^{2} + 28 x - 140$