# How do you determine an equation of a polynomial function with zeros at x=2,-2,1 and y-intercept of 24?

Feb 18, 2016

$y = {x}^{4} + 5 {x}^{3} - 10 {x}^{2} - 20 x + 24$

#### Explanation:

If $\left\{2 , - 2 , 1\right\}$ are all zeros of the polynomial
then
the polynomial must contain the factors:
$\textcolor{w h i t e}{\text{XXX}} \left(x - 2\right) \left(x + 2\right) \left(x - 1\right)$

So
$\textcolor{w h i t e}{\text{XXX}} y = \left(x - 2\right) \left(x + 2\right) \left(x - 1\right) \times a$ for some additional factor $a$

$\overline{y} = \left(x - 2\right) \left(x + 2\right) \left(x - 1\right) = {x}^{3} - {x}^{2} - 4 x + 4$
which is equal to $4$ when $x = 0$
So if $y = \overline{y} \times a$ is to be equal to $24$ when $x = 0$
then $a$ must have the value $6$ when $x = 0$

The most obvious (but not only) value for $a$ is
$\textcolor{w h i t e}{\text{XXX}} \left(x + 6\right)$
in which case
$\textcolor{w h i t e}{\text{XXX}} y = \left(x - 2\right) \left(x + 2\right) \left(x - 1\right) \left(x + 6\right)$
$\textcolor{w h i t e}{\text{XXXX}} = {x}^{4} + 5 {x}^{3} - 10 {x}^{2} - 20 x + 24$