Question

Simplify. Assume that no denominator equals zero.

Answer 1

Before we solve this equation, we need to simplify it. x²-25 will simplify to (x-5)(x+5). x²-2x would simplify to x(x-2) after we factor out a x. x-2 is prime, so we won't do anything to it. x²+5x will simplify to x(x+5) after we factor out an x.

Now this is what we have:

`[(x-5)(x+5) ]/x(x-2) * (x-2)/[x(x+5)]`
If we look at the top and the bottom of the fractions, we see that we have two x-2's and two x+5. Those will cancel out.

`[(x-5)/x]*(1/x)`
Now we will multiply across to get

`(x-5)/x^2`

Answer 2

`(x^2-25)/(x^2-2x)* (x-2)/(x^2+5x)`

Let's factor all four of these before we multiply.

x² - 25
We want two numbers that add to -25 and multiply to 0.
= (x+5)(x-5)

x² - 2x
Both terms are divisible by x.
= x(x-2)

x - 2
unfactorable

x² + 5x
Both terms are divisible by x.
= x(x+5)

Now we have this:


`((x+5)(x-5))/(x(x-2))* (x-2)/(x(x+5))`

Let's go ahead and multiply across.


`((x+5)(x-5)(x-2))/(x(x-2)x(x+5))`

Cancel out the (x+5) and (x-2) from the top and the bottom.


`\boxed{(x-5)/(x^2)}`