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

Sep 29, 2016

$f \left(x\right) = {x}^{4} + 6 {x}^{3} + 14 {x}^{2} + 14 x + 5$

#### Explanation:

Zeros:
$- 1$ multiplicity 2
$\left(- 2 - i\right)$ multiplicity 1

If the zero is $- 1$, the factor is $\left(x - \left(- 1\right)\right) = \left(x + 1\right)$.
A multiplicity of $\textcolor{red}{2}$ implies there are $\textcolor{red}{2}$ factors $\left(x + 1\right)$ or $\left(x + 1\right) \left(x + 1\right) = {\left(x + 1\right)}^{\textcolor{red}{2}} = \left({x}^{2} + 2 x + 1\right)$

For the complex zero $\left(- 2 - i\right)$, there will also be a zero at the complex conjugate$\left(- 2 + i\right)$

The factors are $\left(x - \left(- 2 - i\right)\right)$ and $\left(x - \left(- 2 + i\right)\right)$
or
$\left(x + 2 + i\right) \left(x + 2 - i\right)$
$\left({x}^{2} + 2 x - i x + 2 x + 4 - 2 i + i x + 2 i - {i}^{2}\right)$
$\left({x}^{2} + 4 x + 4 - \left(- 1\right)\right)$
$\left({x}^{2} + 4 x + 5\right)$

Multiply all factors to find the polynomial in standard form
$f \left(x\right) = \left({x}^{2} + 2 x + 1\right) \left({x}^{2} + 4 x + 5\right) =$

${x}^{4} + 4 {x}^{3} + 5 {x}^{2}$
$\textcolor{w h i t e}{a a a a} 2 {x}^{3} + 8 {x}^{2} + 10 x$
$\textcolor{w h i t e}{a a a a a a a a a a} {x}^{2} + 4 x + 5 =$

$f \left(x\right) = {x}^{4} + 6 {x}^{3} + 14 {x}^{2} + 14 x + 5$