Question 173649
490x^2 - 640          28x^2 - 95x +72
_________________ . ___________________ = ? 
49x^2 - 112x + 64   56x^3 - 62x^2 - 144 
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Eliminate anything you can.  There's a factor of 2 in the 1st NUM and the 2nd DEN, and a factor of 5 in the 1st NUM.
   5*(49x^2 - 64)       28x^2 - 95x +72
_________________ . ___________________ 
49x^2 - 112x + 64    28x^3 - 31x^2 - 72 
A slight improvement.  The (49x^2-64) in the 1st NUM can be factored.
   5*(7x-8)*(7x+8)       28x^2 - 95x +72
_________________ . ___________________ 
49x^2 - 112x + 64    28x^3 - 31x^2 - 72 
Ooooh, the 1st DEN can be factored too.
   5*(7x-8)*(7x+8)      28x^2 - 95x +72
_________________ . ___________________ 
   (7x - 8)^2         28x^3 - 31x^2 - 72 
Check to see if the 2 remaining trinomials are divisible by any of the 3 binomials.  It's common in classroom and textbook problems to present problems like that.
Surprise, surprise!!! (not really) 28x^2 - 95x +72 = (7x-8)*(4x-9)
  5*(7x-8)*(7x+8)       (7x-8)*(4x-9)
_________________ . ___________________ 
   (7x - 8)^2         28x^3 - 31x^2 - 72
So the (7x-8) cancels.
   5*(7x-8)*(7x+8)          (4x-9)
_________________ . ___________________ 
   (7x - 8)         28x^3 - 31x^2 - 72
Then the other one cancels.
   5*(7x+8)          (4x-9)
_________________ . ___________________ 
       1             28x^3 - 31x^2 - 72

   5*(7x+8)(4x-9)
___________________
 28x^3 - 31x^2 - 72
It's looking better.

     5*(28x^2 - 31x - 72)
 =   ___________________
     28x^3 - 31x^2 - 72

It's possible you made a typo, and the trinomials should cancel??