Verified Answer :

JacobtheGreat

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
\((x-5)/x^2\)

RedRicin

\(\frac{x^2-25}{x^2-2x}\times \frac{x-2}{x^2+5x}\)

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:

\(\frac{(x+5)(x-5)}{x(x-2)}\times \frac{x-2}{x(x+5)}\)

Let's go ahead and multiply across.

\(\frac{(x+5)(x-5)(x-2)}{x(x-2)x(x+5)}\)

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

\(\boxed{\frac{x-5}{x^2}}\)



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