# What is the standard form of y= (x-3)(x^3-5)*3x^4-5?

Oct 17, 2017

In standard form
$\textcolor{w h i t e}{\text{XXX}} y = 3 {x}^{8} - 9 {x}^{7} - 15 {x}^{5} + 45 {x}^{4} - 5$

#### Explanation:

$\underbrace{\left(x - 3\right) \left({x}^{3} - 5\right)} \cdot 3 {x}^{4} - 5$

$= \underbrace{\left({x}^{4} - 5 x - 3 {x}^{3} + 15\right) \cdot \left(3 {x}^{4}\right)} - 5$

$= \underbrace{\left(3 {x}^{8} - 15 {x}^{5} - 9 {x}^{7} + 45 {x}^{4}\right) - 5}$

$= 3 {x}^{8} - 15 {x}^{5} - 9 {x}^{7} + 45 {x}^{4} - 5$

To write this in standard form the terms must be arranged in descending degree (where degree is the sum of all variable exponents in the term)
{: (ul("term"),color(white)("xxxx"),ul("degree")), (3x^8,,8), (-15x^5,,5), (-9x^7,,7),(45x^4,,4), (-5,,0) :}

Arranged in decreasing degree order:
$y = 3 {x}^{8} - 9 {x}^{7} - 15 {x}^{5} + 45 {x}^{4} - 5$