SOLUTION: 490x^2 - 640 28x^2 - 95x +72 ___________ _____ . ___________ _______ = ? 49x^2 - 112x + 64 56x^3 - 62x^2 - 144 I do not know how to start this problem. Thank you

SOLUTION: 490x^2 - 640 28x^2 - 95x +72 _________ _ . ___ _ = ? 49x^2 - 112x + 64 56x^3 - 62x^2 - 144 I do not know how to start this problem. Thank you

Question 173649: 490x^2 - 640 28x^2 - 95x +72
_________________ . ___________________ = ?
49x^2 - 112x + 64 56x^3 - 62x^2 - 144

I do not know how to start this problem. Thank you for your help
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
490x^2 - 640 28x^2 - 95x +72
_________________ . ___________________ = ?
49x^2 - 112x + 64 56x^3 - 62x^2 - 144
---------------------
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??

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