The numerator is (48y^4+64y^3-6)/(8y^2)=(24y^4+32y^3-3)/(4y^2)
The denominator is 3y-3/4y^2+4=(12y^3+16y^2-3)/(4y^2)
So we have it= (24y^4+32y^3-3)/(12y^3+16y^2-3) after cancelling the 4y^2
Now you can easily see that it could not be c or d since when multiplying (12y^3+16y^2-3) with c or d (we should get then (24y^4+32y^3-3) ) the free terms (the ones with no power of y) gives +3 for c and -15 for d but it should be -3 - so none is correct. It is also not a since multiplying y+1 by (12y^3+16y^2-3) gives us powers of y^4, y^3, y^2, y - but we don't have y in (12y^3+16y^2-3). It is not a then.
The final result is then (24y^4+32y^3-3)/(12y^3+16y^2-3) and can't simplify it further. (we could if the numerator had a -6 instead of the -3 - it would be simply "2" - but it isn't the way the problem is stated here)
May 29th, 2015
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