# How do you find a polynomial of degree 3 that has zeros of -3, 0, 1?

Nov 15, 2016

$P \left(x\right) = {x}^{3} + 2 {x}^{2} - 3 x$

#### Explanation:

A polynomial has $\alpha$ as a zero if and only if $\left(x - \alpha\right)$ is a factor of the polynomial. Working backwards, then, we can generate a polynomial with any zeros we desire by multiplying such factors.

We want a polynomial $P \left(x\right)$ with zeros $- 3 , 0 , 1$, so:

$P \left(x\right) = \left(x - \left(- 3\right)\right) \left(x - 0\right) \left(x - 1\right)$

$= \left(x + 3\right) x \left(x - 1\right)$

$= x \left(x + 3\right) \left(x - 1\right)$

$= x \left({x}^{2} + 2 x - 3\right)$

$= {x}^{3} + 2 {x}^{2} - 3 x$

Note that we could also multiply by any nonzero constant without changing the zeros, if a different polynomial is desired.