# How do you find a third degree polynomial given roots 3 and 2-i?

Jun 18, 2018

We are given roots

${x}_{1} = 3$
${x}_{2} = 2 - i$

The complex conjugate root theorem states that, if $P$ is a polynomial in one variable and $z = a + b i$ is a root of the polynomial, then $\overline{z} = a - b i$, the conjugate of $z$, is also a root of $P$.

As such, the roots are

${x}_{1} = 3$
${x}_{2} = 2 - i$
${x}_{3} = 2 - \left(- i\right) = 2 + i$

From Vieta's formulas, we know that the polynomial $P$ can be written as:

${P}_{a} \left(x\right) = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

Where $a$ is a constant.

$\therefore {P}_{a} \left(x\right) = a \left(x - 3\right) \left(x - 2 + i\right) \left(x - 2 - i\right)$

$\therefore {P}_{a} \left(x\right) = a \left({x}^{3} - 7 {x}^{2} + 17 x - 15\right)$

One such polynomial would be

${P}_{1} \left(x\right) = {x}^{3} - 7 {x}^{2} + 17 x - 15$