# How do you write a polynomial in standard form given zeros 3, 2, -1?

Mar 5, 2016

${x}^{3} - 4 {x}^{2} + x + 6$
Using the fact that if $\left(x - a\right)$ is a factor of a polynomial, then $a$ is a root of that polynomial, we can generate a polynomial with a given set of roots easily by starting from its factored form.
In this case, given the roots $3$, $2$, and $- 1$, we would have
$P \left(x\right) = \left(x - 3\right) \left(x - 2\right) \left(x - \left(- 1\right)\right) = {x}^{3} - 4 {x}^{2} + x + 6$