# Question #d4dce

Feb 23, 2018

See below.

#### Explanation:

First, get rid of the fraction by multiplying everything by $x$

$2 {x}^{2} + \frac{36000}{x} \implies 2 {x}^{3} + 36000$

Now we take out the $H C F$ of each term, in this case $2$

$2 {x}^{3} + 36000 \implies 2 \left({x}^{3} + 18000\right)$

Feb 24, 2018

$\frac{2 \left({x}^{3} + 18000\right)}{x}$

#### Explanation:

Let's set $y$ to equal $2 {x}^{2} + \frac{36000}{x}$

We have:

$y = 2 {x}^{2} + \frac{36000}{x}$ Multiply both sides by $x$.

$x y = x \left(2 {x}^{2} + \frac{36000}{x}\right)$Distribute

$x y = 2 {x}^{3} + 36000$

$x y = 2 \cdot x \cdot x \cdot x + 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5$

We ask ourselves: What do $2 \cdot x \cdot x \cdot x$ and $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5$ share in common?

We see that they share a $2$

We factor. (Take out a two from each term)

$x y = 2 \left(x \cdot x \cdot x + 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5\right)$

$x y = 2 \left({x}^{3} + 18000\right)$

We now divide both sides by $x$.

$y = \frac{2 \left({x}^{3} + 18000\right)}{x}$

This is equal to what we started with ($2 {x}^{2} + \frac{36000}{x}$) because they both equal $y$.

The answer is $\frac{2 \left({x}^{3} + 18000\right)}{x}$