# How do you write a polynomial in standard form given zeros 3, -4, 2?

${x}^{3} - {x}^{2} - 14 x + 24 = 0$

#### Explanation:

The method that I am used to is by equating each zero to the variable x

that is

$x = 3$ and $x = - 4$ and $x = 2$

transposing the numbers to the left side, we have

$x - 3 = 0$ and $x + 4 = 0$ and $x - 2 = 0$

so that we have the factors

$\left(x - 3\right) \cdot \left(x + 4\right) \cdot \left(x - 2\right) = 0$

multiply them to obtain the standard form

$\left({x}^{2} + x - 12\right) \left(x - 2\right) = 0$
$\left({x}^{3} + {x}^{2} - 12 x - 2 {x}^{2} - 2 x + 24\right) = 0$
simplify to obtain the final answer

${x}^{3} - {x}^{2} - 14 x + 24 = 0$

God bless.... I hope the explanation is useful.