Fraction Help !?

Question

I have to write this equation in simplest form and all exponents have to be positive

Here it is ....

48a^5b
----------
12ab^2

Also I need help with the division of fractions like this one

X^2 +2x-15
---------------
4x^2 divided by

x^2-25
----------
2x-10

HELP ! :)

Answer 1

For the first one, just cancel out anything that cancels in the top and bottom. 48 = 12*4, so the 12's cancel and you have 4 left in the top.
a^5 = a*a*a*a*a, so one of the a's cancels with the one in the bottom, which leaves a^4 on top.
b^2 = b*b, so one of the b's cancels and there is 1 b left on the bottom, so you are left with 4a^4/b

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The first thing you need to do on this one is rewrite it as a multiplication problem. Flip the bottom fraction and multiply with the first:

(x^2 + 2x - 15) * (2x-10)
-------------------------------
4x^2 * (x^2 - 25)

Now factor as much as you can

x^2 + 2x - 15 = (x-3)(x+5)
2x-10 = 2(x-5)
x^2 - 25 = (x+5)(x-5)

So the new expression is

(x-3)(x+5)(2)(x-5)
-----------------------
4x^2(x+5)(x-5)

Cancel the (x+5) and the (x-5) and the 2 with the 4 and you're left with

(x-3)/2x^2

Answer 2

48a^5b
---------- = (4a^4)/b
12ab^2

Also I need help with the division of fractions like this one

X^2 +2x-15
---------------
4x^2 divided by

x^2-25
----------
2x-10

x^2 +2x-15 (2X- 10 )
= --------------- times -------------------
4x^2 x^2-25

=(x+5)(x-3)/(4x^2) * 2(x-5)/[(x+5)(x-5)]
[2(x-5)(x+5)(x-3)]/[4x^2(x+5)(x-5)]
=(x-3)/2x^2

Answer 3

4a^4 all over b. As far as the next one, just flip the second equation and multiply across.