# How do you write a polynomial in standard form given zeros 1 (multiplicity 2), -2 (multiplicity 3)?

May 26, 2016

${x}^{5} + 4 {x}^{4} + {x}^{3} - 10 {x}^{2} - 4 x + 8 = 0$

#### Explanation:

If $\left\{\alpha , \beta , \gamma , \delta , . .\right\}$ are the zeros of a function, then the function is

$\left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right) \left(x - \delta\right) \ldots = 0$

Here zeros are $1$ (multiplicity $2$) and $- 2$ (multiplicity $3$), hence function is

$\left(x - 1\right) \left(x - 1\right) \left(x + 2\right) \left(x + 2\right) \left(x + 2\right) = 0$ or

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

$\left({x}^{2} - 2 x + 1\right) \left({x}^{3} + 6 {x}^{2} + 12 x + 8\right) = 0$ or

${x}^{2} \left({x}^{3} + 6 {x}^{2} + 12 x + 8\right) - 2 x \left({x}^{3} + 6 {x}^{2} + 12 x + 8\right) + 1 \left({x}^{3} + 6 {x}^{2} + 12 x + 8\right) = 0$

${x}^{5} + 6 {x}^{4} + 12 {x}^{3} + 8 {x}^{2} - 2 {x}^{4} - 12 {x}^{3} - 24 {x}^{2} - 16 x + {x}^{3} + 6 {x}^{2} + 12 x + 8 = 0$

${x}^{5} + 4 {x}^{4} + {x}^{3} - 10 {x}^{2} - 4 x + 8 = 0$