# How do you find a polynomial function of degree 6 with -1 as a zero of multiplicity 3, 0 as a zero of multiplicity 2, and 1 as a zero of multiplicity 1?

Nov 5, 2016

$f \left(x\right) = {x}^{6} + 2 {x}^{5} - 2 {x}^{3} - {x}^{2}$

#### Explanation:

Each zero (e.g. $a$) corresponds to a linear factor (e.g. $\left(x - a\right)$).

Multiplicity corresponds to a repetition of that factor.

So in our example, the following polynomial fits the criteria:

$f \left(x\right) = {\left(x - \left(- 1\right)\right)}^{3} {\left(x - 0\right)}^{2} \left(x - 1\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {\left(x + 1\right)}^{3} {x}^{2} \left(x - 1\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{2} {\left(x + 1\right)}^{2} \left(x - 1\right) \left(x + 1\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{2} \left({x}^{2} + 2 x + 1\right) \left({x}^{2} - 1\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{2} \left({x}^{4} + 2 {x}^{3} - 2 x - 1\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{6} + 2 {x}^{5} - 2 {x}^{3} - {x}^{2}$

Any polynomial in $x$ with these zeros in these multiplicities will be a multiple (scalar or polynomial) of this $f \left(x\right)$.