# How do you write a polynomial in standard form given the zeros (5 ,+i√3 , -i√3)?

Mar 25, 2016

f(x))=x^3-5x^2+9x-45

#### Explanation:

A polynomial with zeros $\left\{a , b , c\right\}$ is given by $f \left(x\right) = \left(x - a\right) \left(x - b\right) \left(x - c\right)$.

As the zeros are $\left\{5 , + i \sqrt{3} , - i \sqrt{3}\right\}$,

$f \left(x\right) = \left(x - 5\right) \left(x - i \sqrt{3}\right) \left(x - \left(- i \sqrt{3}\right)\right) = \left(x - 5\right) \left(x - i \sqrt{3}\right) \left(x + i \sqrt{3}\right)$ or

$f \left(x\right) = \left(x - 5\right) \left({x}^{2} - {\left(i \sqrt{3}\right)}^{2}\right)$ or

$f \left(x\right) = \left(x - 5\right) \left({x}^{2} - \left(- 9\right)\right) = \left(x - 5\right) \left({x}^{2} + 9\right)$ or

$f \left(x\right) = x \left({x}^{2} + 9\right) - 5 \left({x}^{2} + 9\right) = {x}^{3} + 9 x - 5 {x}^{2} - 45$ or

f(x))=x^3-5x^2+9x-45