# How do you find a third degree polynomial given roots -1 and 3+i?

Oct 20, 2016

The third-degree polynomial with roots $- 1$ and $3 + i$ is
$f \left(x\right) = {x}^{3} - 5 {x}^{2} + 4 x + 10$.

#### Explanation:

Complex roots are always in pairs, and the pairs are conjugates. Therefore, even though you are only given two roots, the polynomial actually has three roots. The third root is $3 - i$.

Remember that a root is represented by $k$, and that the factor which yields a root is in the form $x - k$. Therefore, to write the polynomial which has the given roots and a leading coefficient of $1$, simply set up the roots in factor form and multiply them.

$f \left(x\right) = \left(x - - 1\right) \left(x - \left[3 + i\right]\right) \left(x - \left[3 - i\right]\right)$
$f \left(x\right) = \left(x + 1\right) \left(x - 3 - i\right) \left(x - 3 + i\right)$
$f \left(x\right) = \left(x + 1\right) \left({x}^{2} - 3 x + i x - 3 x + 9 - 3 i - i x + 3 i - {i}^{2}\right)$
$f \left(x\right) = \left(x + 1\right) \left({x}^{2} - 6 x + 9 + 1\right)$
$f \left(x\right) = \left(x + 1\right) \left({x}^{2} - 6 x + 10\right)$
$f \left(x\right) = {x}^{3} - 6 {x}^{2} + 10 x + {x}^{2} - 6 x + 10$
$f \left(x\right) = {x}^{3} - 5 {x}^{2} + 4 x + 10$