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# Dividing fractions that contain exponents?

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.

(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)

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

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.