Dividing fractions that contain exponents?

Question

I can't seem to figure this one out.  It's more complicated than others I've been doing.  Thank you for your help and explanation!

 

27n²-147

3n²-16n+21

Divided by...

15n²+32n-7

n²-9n+18

 

I know the second fraction needs to be flipped for multiplication so it would become:

n²-9n+18

15n²+32n-7

Next is factoring but I can't seem to figure it.

Answer 1

(27n² -147) / (3n2 - 16 + 21) * (n² -9n +18) / (15n² + 32n - 7)

equivalent to:

(3n + 7) (9n - 21) / (3n -7) (n - 3) * (n - 6) (n - 3) / (3n + 7) (5n - 1)

3 (3n - 7) / (3n - 7) * (n - 6) /( 5n - 1)

therefore:

3 (n - 6) / (5n - 1)

 

 

 

Answer 2

27n² - 147 = 3(9n² - 49) = 3(3n - 7)(3n + 7) <--- numerator
3n² -16n + 21 = (3n - 7)(n - 3) <--- denominator , there is common factor (3n - 7), let's divide both numerator and denominator by (3n - 7) ----->
     3(3n + 7)    
       n - 3
Now, let's simplify the divider of the original rational expression:
15n² + 32n - 7 = (3n + 7)(5n - 1)
n² - 9n + 18 = (n - 6)(n - 3)
    (3n + 7)(5n - 1)  
     (n - 6)(n - 3)  
             
      3(3n + 7)   (n - 6)(n - 3)                
       (n - 3)    (3n + 7)(5n - 1)             

              3(n - 6)        or      3n - 18     
               5n - 1                    5n - 1  

Answer 3

That fact that there are only 2 term in 27n²-147 may suggest it is a perfect square. And after dividing by 3 it is! 3(9n²-49). Factor each quadratic and cancel and you will find the answer.