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

Question

1345 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-17T14:21:38

See below.

We have that

#x^2=q(x)(2x+4)+r(x)#

where #q(x)# is the quocient, and #r(x)# is the remainder

Analyzing the degrees involved we have:

#q(x) = ax+b#

#r(x) = c#

so

#x^2=2ax^2+2(b+2a)x+4b+c#

so we need

#{(2a=1),(b+2a=0),(4b+c=0):}#

Solving for #a,b,c# we have #a=1/2,b=-1,c=4#

and finally

#q(x)=x/2-1#

#r(x)=4#