# What is the standard form of y= (4x-2x^2)(-4x+3) -(x-4)^3?

Jun 23, 2018

$7 {x}^{3} - 10 {x}^{2} - 36 x + 64$

#### Explanation:

In order to complete this problem (and make it easier on ourselves), we should split it into two parts. Let's start with the first part, $\left(4 x - 2 {x}^{2}\right) \left(- 4 x + 3\right)$. Using FOIL (First, Outsides, Insides, Last), we can simplify this down:

$\left(4 x\right) \left(- 4 x\right) + \left(4 x\right) \left(3\right) + \left(- 2 {x}^{2}\right) \left(- 4 x\right) + \left(- 2 {x}^{2}\right) \left(3\right)$
$- 16 {x}^{2} + 12 x + 8 {x}^{3} - 6 {x}^{2}$

We don't need to write this in standard form quite yet as we still have the other half of the equation, ${\left(x - 4\right)}^{3}$. The formula for the multiplication of ${\left(a - b\right)}^{3}$ is ${a}^{3} + 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$. Let's use this as a guideline for our multiplication.

${\left(x - 4\right)}^{3}$
${x}^{3} + \left(3\right) \left({x}^{2}\right) \left(- 4\right) + \left(3\right) \left(x\right) {\left(- 4\right)}^{2} + {\left(- 4\right)}^{3}$
${x}^{3} - 12 {x}^{2} + 48 x - 64$

Let's now combine our two parts to create one answer:

$- 16 {x}^{2} + 12 x + 8 {x}^{3} - 6 {x}^{2} - \left({x}^{3} - 12 {x}^{2} + 48 x - 64\right)$
$- 16 {x}^{2} + 12 x + 8 {x}^{3} - 6 {x}^{2} - {x}^{3} + 12 {x}^{2} - 48 x + 64$
$8 {x}^{3} - {x}^{3} - 16 {x}^{2} - 6 {x}^{2} + 12 {x}^{2} + 12 x - 48 x + 64$
$7 {x}^{3} - 10 {x}^{2} - 36 x + 64$