Posted by **jasort20** on Saturday, March 17, 2007 at 1:51am.

Can someone show me how do i multiply the following fractions:

(n^2-n-20)/(2n^2) times (n^2+5n)/(n^2-25)

When you multiply fractions, you just multiply straight across. So if you have:

1/3 * 1/4

1*1

----

3*4

That's 1/12

With that in mind, let's look at your question:

(n^2-n-20)(n^2+5n)

-------------------

(2n^2)(n^2-25)

Let's start off, to make it possibly a little easier, by factoring out what we can:

(n^2-n-20) can be written:

(n-5)(n+4)

So now we have

(n-5)(n+4)(n^2+5n)

------------------

(2n^2)(n^2-25)

The n^2+5n can be:

n(n+5)

Phew. So...all together on top, we have

(n-5)(n+4)(n)(n+5)

Now, the bottom.

2n^2 can be left alone.

n^2-25 can be:

(n-5)(n+5)

So now we have, on the bottom:

(2n^2)(n-5)(n+5)

All together, we have:

(n-5)(n+4)(n)(n+5)

-------------------

(2n^2)(n-5)(n+5)

The (n+5)s and the (n-5)s cancel each other out. So we're left with:

(n+4)(n)

--------

2n^2

That's:

n^2 + 4n

--------

2n^2

Can factor n out :

n(n+4)

-----

n(2n)

You're left with:

(n+4) / (2n)

I put a lot in here. So feel free to ask specific questions. Maybe someone else on the board can explain without as much detail.

Wow let me just say EXTRA THANK YOU. the way how you detailed it, it is very good yes its a lot but well understandable i can see what you mean what to do. Gee this is a better explanation than my own instructor or even the book......thank you....

I'm glad I could help. To be honest, I don't figure out a lot of these math problems before I post them. I just start typing and explain what I'm doing as I go along. I have, in the past, had to erase entire posts and start over when I realized I made a mistake in the 2nd step. LOL

Glad this answer helped.

Matt

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