SOLUTION: use long division to solve (2x-5)/(x^3-5x^2+5x+4) I know how to do the long division method but the leading 2 in 2x-5 is throwing me off.

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Question 403610: use long division to solve (2x-5)/(x^3-5x^2+5x+4)
I know how to do the long division method but the leading 2 in 2x-5 is throwing me off.
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website!
use long division to solve (2x-5)/(x^3-5x^2+5x+4)
I know how to do the long division method but the leading 2 in 2x-5 is throwing me off.
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The problem here is that you are used to multipliers of whole numbers. For the given problem, your multipliers in the quotient are fractions. For example the multiplier in the quotient for the first term is (1/2)x^2, for the second term, (-5/4)x, and for the third term, 5/8. If I did it right, there was a remainder of 7/8.
The quotient I got was (1/2)x^2-(5/4)x+(5/8)+remainder (7/8)/(2x-5)
This is a strange problem and it would be near impossible to find the zeros of tlhis function. In fact there is only one real zero as seen on the graph of this function below.
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