# How do you create a polynomial p which has zeros c=1, c=3, c=-3 is a zero of multiplicity 2, the leading term is -5x^3?

Aug 6, 2017

$- 5 {x}^{3} + 5 {x}^{2} + 45 x - 45$

#### Explanation:

Note that $x$ is a zero of a polynomial if and only if $\left(x - c\right)$ is a factor of that polynomial.

So in order to have zeros $1$, $3$ and $- 3$, our polynomial must be a multiple of:

$\left(x - 1\right) \left(x - 3\right) \left(x + 3\right) = \left(x - 1\right) \left({x}^{2} - 9\right) = {x}^{3} - {x}^{2} - 9 x + 9$

In order that its leading term be $- 5 {x}^{3}$, we just need to multiply it by $- 5$ to get:

$- 5 \left({x}^{3} - {x}^{2} - 9 x + 9\right) = - 5 {x}^{3} + 5 {x}^{2} + 45 x - 45$