How do you find the quotient of #(2y^2-3y+1)div(y-2)# using long division?

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Answer 1 · 2017-05-31T09:13:18

The quotient is #=(2y+1)# and the remainder is #=3#

Let's perform the long division

#color(white)(aaaa)##y-2##|##color(white)(aaaa)##2y^2-3y+1##|##color(white)(aa)##2y+1#

#color(white)(aaaaaaaaaaaaaa)##2y^2-4y#

#color(white)(aaaaaaaaaaaaaaaa)##0+y+1#

#color(white)(aaaaaaaaaaaaaaaaaaa)##y-2#

#color(white)(aaaaaaaaaaaaaaaaaaa)##0+3#

So,

#(2y^2-3y+1)/(y-2)=(2y+1)+3/(y-2)#

The quotient is #=(2y+1)# and the remainder is #=3#

Answer 2 · 2017-05-31T14:34:55

Quotient #= (2y+1)#

I know the question asked for long division, but there is an easier way to do it.

We want #(2y^2-3y+1)/(y-2)#

First, let #f(y)=2y^2-3y+1#

Then take #f(2)=2(2^2)-3(2)+1=3#

So #(2y^2-3y+1)/(y-2)# has a remainder of #3#.

Therefore #2y^2-3y+1=(ay+b)(y-2)+3#

#2y^2-3y-2=(ay+b)(y-2)#

#y^2# has a coefficient of #2#, so #ay^2=2y^2 rArr a=2#

The constant is #-2#, so #-2b=-2rArrb=1#

Thus the quotient, #(ay+b)#, equals #(2y+1)#