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

Dec 27, 2015

$y = 4 {x}^{3} + 16 {x}^{2} - 40 x + 20$

#### Explanation:

For convenience, separate out the scalar factor $4$ temporarily while multiplying out, group the terms in descending degree and combine. For illustration I have shown more steps than normal:

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

$= 4 \left(x - 1\right) \left({x}^{2} + 5 x - 5\right)$

$= 4 \left(x \left({x}^{2} + 5 x - 5\right) - 1 \left({x}^{2} + 5 x - 5\right)\right)$

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

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

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

$= 4 \left({x}^{3} + \left(5 - 1\right) {x}^{2} + \left(- 5 - 5\right) x + 5\right)$

$= 4 \left({x}^{3} + 4 {x}^{2} - 10 x + 5\right)$

$= 4 {x}^{3} + 16 {x}^{2} - 40 x + 20$

Alternatively, just look at the combinations of terms to give each power of $x$ in descending order like this:

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

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

$= 4 {x}^{3} + 16 {x}^{2} - 40 x + 20$