# Question #a1a2d

Jun 13, 2017

${x}^{4} - 10 {x}^{3} + 20 {x}^{2} + 40 x - 96$

#### Explanation:

If we already have the zeros they original came from solving for x:
This is what we have right now $x = - 2 , 2 , 4 , 6$
So we work our way backwards:

$x = - 2$ You add two to both sides
$x + 2 = \cancel{- 2 + 2}$
$\textcolor{b l u e}{x + 2}$ This is one factor

$x = 2$ Subtract two from both sides
$x - 2 = \cancel{2 - 2}$
$\textcolor{b l u e}{x - 2}$ Another factor

$x = 4$ Subtract four from both sides
$x - 4 = \cancel{4 - 4}$
$\textcolor{b l u e}{x - 4}$ Third factor

$x = 6$ Subtract six from both sides
$x - 6 = \cancel{6 - 6}$
$\textcolor{b l u e}{x - 6}$ Final factor

Now that we know this information we can now multiply out our factors of:

$\left(x + 2\right) \left(x - 2\right) \left(x - 4\right) \left(x - 6\right)$

$\left({x}^{2} - 4\right) \left(x - 4\right) \left(x - 6\right)$

$\left({x}^{3} - 4 {x}^{2} - 4 x + 16\right) \left(x - 6\right)$

${x}^{4} - 6 {x}^{3} - 4 {x}^{3} + 24 {x}^{2} - 4 {x}^{2} + 24 x + 16 x - 96$

Simplify:

${x}^{4} - 10 {x}^{3} + 20 {x}^{2} + 40 x - 96$