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.
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