`(2x^2-18)/(x^3-25x)xx(3x-15)/(2x^2+6x)`

We need to factor everything on the top and bottom of the fractions.

For the top of the first fraction:

`2x^2− 18 = 2(x^2− 9)`

We recognise the expression in brackets to be a difference of 2 squares, which we learned about before. We can write:

`x^2− 9 = (x + 3)(x − 3)`

The bottom of the first fraction also uses difference of 2 squares:

`x^3− 25x` ` = x(x^2− 25)` ` = x(x + 5)(x −5)`

The top of the second fraction requires taking the common factor of 3 outside:

`3x − 15 = 3(x − 5)`

The bottom of the second fraction has 2x as a common factor :

`2x^2+ 6x = 2x(x + 3)`

So far we've done:

`(2x^2-18)/(x^3-25x)xx(3x-15)/(2x^2+6x) `

`=(2(x^2-9))/(x(x^2-25))xx(3(x-5))/(2x(x+3) `

`= (2(x+3)(x-3))/(x(x+5)(x-5))xx(3(x-5))/(2x(x+3)`

Next we cancel out each of the following from top and bottom:

`2`,
` (x + 3)`, and
`(x − 5)`

This gives:

`=(x-3)/(x(x+5))xx3/x`

The final step is to multiply tops and multiply bottoms, since we cannot cancel anything else.

`=(3(x-3))/(x^2(x+5))`

So

`(2x^2-18)/(x^3-25x)xx(3x-15)/(2x^2+6x)=(3(x-3))/(x^2(x+5))`