# How do you find a polynomial function of degree 4 with -2 as a zero of multiplicity 3 and 0 as a zero of multiplicity 1?

Oct 19, 2015

${x}^{4} + 6 {x}^{3} + 12 {x}^{2} + 8 x$

#### Explanation:

${y}_{a} = \left(x + 2\right) \left(x + 2\right) \left(x + 2\right) = {x}^{3} + 6 {x}^{2} + 12 x + 8$
has $\left(- 2\right)$ as its only root (but with multiplicity of $3$)

${y}_{b} = x$
has $\left(0\right)$ as its only root (multiplicity of $1$)

$y = {y}_{a} \cdot {y}_{b} = {x}^{4} + 6 {x}^{3} + 12 {x}^{2} + 8 x$
has $\left(- 2\right)$ as a root (multiplicity $3$) and $\left(0\right)$ as a root (multiplicity (1))
and is of degree $4$