Help with adding/subtracting fractions with polynomials on the bottom

[NYKn1cks11]

The problem:

(3)/(9m^2-48m+64)-(m)/(64-9m^2)

I know the answer is 3m^2+m+24/(3m-8)^2(3m+8) but i don't know how to get it.

 

[procyon]

look for \(\displaystyle 3m-8\) in the denominators
also note that \(\displaystyle (a+b)(a-b) = a^2-b^2\)

 

[Deleted member 4993]

The problem:

(3)/(9m^2-48m+64) -(m)/(64-9m^2)

I know the answer is 3m^2+m+24/(3m-8)^2(3m+8) but i don't know how to get it.

denominator 1: .... 9m^2-48m+64 .... factorize it

denominator 2: .... 64-9m^2 .... factorize it

Find LCM

Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[NYKn1cks11]

The factoring part is easy but what do you do once you get every term in it's simplest form?Like how do you know what to multiply the numerators by?

 

[procyon]

What have you got now that you have factorized it?

 

[NYKn1cks11]

I factored it and got:

_____3_____ - ______m_____
(3m+8)(3m-8) (8+3m)(8-3m)


How do you find the LCD from here?

 

[procyon]

The correct factorization is

\(\displaystyle \dfrac{3}{(3m-8)^2} -\dfrac{m}{(8-3m)(8+3m)} \)

On the lhs of the minus you could also have written

\(\displaystyle \dfrac{3}{64-48m+9m^2}\)

which you would probably have factorized as

\(\displaystyle \dfrac{3}{(8-3m)^2}\)