How do you divide # (7x-5x^2 – 3+6x^2) /( 2x-1)#?

1 Answer
Jul 26, 2017

#x/2 + 15/4 + 3/(4(2x-1))#

Explanation:

Given: #(7x-5x^2 - 3 + 6x^2)/(2x-1)#

First add like terms in the numerator: #(x^2+7x-3)/(2x-1)#

Long division:

step 1. #"What" xx 2x = x^2?: " "1/2x#

#" "1/2 x#
#" "2x-1 |bar(x^2 + 7x - 3)#
#" "ul(x^2 - 1/2x)#

#" "#To subtract, change the #7# to #7/1 xx 2/2 = 14/2#:

#" "1/2 x#
#" "2x-1 |bar(x^2 + 14/2x - 3)#
#" "ul(x^2 - 1/2x)#
#" " 15/2x - 3#

step 2. #"What" xx 2x = 15/2x?: " "(15x)/4 xx (2x)/1 = (30x)/4 = (15x)/2#

#" "1/2 x + 15/4#
#" "2x-1 |bar(x^2 + 14/2x - 3)#
#" "ul(x^2 - 1/2x)#
#" " 15/2x - 3#
#" "ul(15/2x - 15/4)#

#" "#To subtract, change the #-3# to #-3/1 xx 4/4 = -12/4#:

#" "1/2 x + 15/4#
#" "2x-1 |bar(x^2 + 14/2x - 3)#
#" "ul(x^2 - 1/2x)#
#" " 15/2x - 12/4#
#" "ul(15/2x - 15/4)#
#" "3/4#

The remainder #= 3/4/ (2x-1) = 3/(4(2x-1))#

So, #(x^2+7x-3)/(2x-1) = x/2 + 15/4 + 3/(4(2x-1))#