# How do you find a polynomial function that has zeros x=-5, 1, 2 and degree n=4?

Aug 10, 2018

$f \left(x\right) = {\left(x + 5\right)}^{2} \left(x - 1\right) \left(x - 2\right) = {x}^{4} + 7 {x}^{3} - 3 {x}^{2} - 55 x + 50$

#### Explanation:

If a polynomial function $f \left(x\right)$ has zeros $x = - 5$, $x = 1$ and $x = 2$, then it has factors $\left(x + 5\right)$, $\left(x - 1\right)$ and $\left(x - 2\right)$.

If these were its only factors, then it would be a cubic.

It is not clear from the question whether $- 5$, $1$ and $2$ are supposed to be the only zeros. If so then one of them must be of multiplicity $2$.

In any case, a suitable quartic would be:

$f \left(x\right) = {\left(x + 5\right)}^{2} \left(x - 1\right) \left(x - 2\right) = {x}^{4} + 7 {x}^{3} - 3 {x}^{2} - 55 x + 50$