How do you divide #(2x^2-2x+4)/( 2x-1)#?

1 Answer
Haunt
Nov 5, 2015

#x-(3/2)+5/(4x-2)#

Explanation:

Polynomial division works much the same way as regular long division:

Find a value that multiplied to the first term of the denominator is equal to the first term of the numerator (sorry, formatting this is a little hard)
#(2x-1)*x=2x^2-x#

so your first term is x

next subtract the numerator by new value
#(2x^2-2x+4)-(2x^2-x)=-3x+4#

Then repeat the first term for this new value, find a value that multiplied by your (original) denominator is equal to the new value
#(2x-1)*-3/2=-3x+(3/2)#

so your second term is #-(3/2)#

subtract the numerator (from after the first subtraction) by this new value
#(-3x+4)-(-3x+(3/2))=5/2#

Finally like the remainder of normal long division you make a fraction over the divisor
#(5/2)/(2x-1)=5/(4x-2)#

This is your last term leaving you with
#x-(3/2)+5/(4x-2)#

A great resource for these sorts of problems if you can't understand my formatting: http://www.purplemath.com/modules/polydiv2.htm

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