# How do you form a polynomial function whose zeros, multiplicities and degrees are given: Zeros: -2, multiplicity 2; 4, multiplicity 1; degree 3?

Jul 1, 2016

${x}^{3} - 12 x - 16$

#### Explanation:

the polynomial will be composed by the product of 3 (due to the degree) bynomials with degree 1:

${\left(x + 2\right)}^{2}$ that is null if x=-2, multiplicity 2,

and

x-4 that is null if x=4, so the polynomial is obtained by multiplyng:

${\left(x + 2\right)}^{2} \left(x - 4\right)$

$\left({x}^{2} + 4 x + 4\right) \left(x - 4\right)$

${x}^{3} \cancel{- 4 {x}^{2}} \cancel{+ 4 {x}^{2}} - 16 x + 4 x - 16$

that's

${x}^{3} - 12 x - 16$