# What is the standard form of y= (x^2-4)(2x-4) -(2x+5)^2?

Dec 27, 2017

$y = 2 {x}^{3} - 8 {x}^{2} - 28 x - 9$

#### Explanation:

$y = \left({x}^{2} - 4\right) \left(2 x - 4\right) - {\left(2 x + 5\right)}^{2}$

$y = \left(\left({x}^{2}\right) \left(2 x\right) + \left(- 4\right) \left(2 x\right) + \left({x}^{2}\right) \left(- 4\right) + \left(- 4\right) \left(- 4\right)\right) - \left({\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(5\right) + {\left(5\right)}^{2}\right)$

$y = 2 {x}^{3} - 8 x - 4 {x}^{2} + 16 - 4 {x}^{2} - 20 x - 25$

$y = 2 {x}^{3} - 8 {x}^{2} - 28 x - 9$

Dec 27, 2017

$y = 2 {x}^{3} - 8 {x}^{2} - 28 x - 9$

#### Explanation:

We have to expand a bunch:

$\left({x}^{2} - 4\right) \left(2 x - 4\right) = 2 {x}^{3} - 4 {x}^{2} - 8 x + 16$
${\left(2 x + 5\right)}^{2} = 4 {x}^{2} + 20 x + 25$

$\left({x}^{2} - 4\right) \left(2 x - 4\right) - {\left(2 x + 5\right)}^{2}$

Replace with expansions:
$= 2 {x}^{3} - 4 {x}^{2} - 8 x + 16 - \left(4 {x}^{2} + 20 x + 25\right)$

Distribute the negative:
$= 2 {x}^{3} - 4 {x}^{2} - 8 x + 16 - 4 {x}^{2} - 20 x - 25$

Collect like terms:
$= 2 {x}^{3} - 8 {x}^{2} - 28 x - 9$

So our final answer is $y = 2 {x}^{3} - 8 {x}^{2} - 28 x - 9$