How can I change #f(x) = (2x-1)/(x+2)# into the form #A + B/(x+2)# where A and B are both integers?

1 Answer
Aug 20, 2017

#2-5/(x+2)#

Explanation:

#"using "(x+2)" as a factor in the numerator"#

#"consider the numerator"#

#2(x+2)color(magenta)(-4)-1#

#=2(x+2)-5#

#rArr(2(x+2)-5)/(x+2)=2-5/(x+2)#

#color(blue)"As a check"#

#2-5/(x+2)#

#=(2(x+2))/(x+2)-5/(x+2)#

#=(2x+4-5)/(x+2)=(2x-1)/(x+2)larr" True"#