How do you express #(x²+2) / (x+3) # in partial fractions?

1 Answer
Maggie
Feb 22, 2016

#x/1 + {-3x+2}/{x+3}#

Explanation:

because the top quadratic and the bottom is linear you're looking for something or the form

#A/1 + B/(x+3)#, were #A# and #B# will both be linear functions of #x# (like 2x+4 or similar).

We know one bottom must be one because x+3 is linear.

We're starting with
#A/1 + B/(x+3)#.
We then apply standard fraction addition rules. We need to get then to a common base.

This is just like numerical fractions #1/3+1/4=3/12+4/12=7/12.#

#A/1 + B/(x+3)=> {A*(x+3)}/{1*(x+3)} + B/(x+3)={A*(x+3)+B}/{x+3}#.
So we get the bottom automatically.

Now we set #A*(x+3)+B=x^2+2#
#Ax + 3A + B=x^2+2#
#A# and #B# are linear terms so the #x^2# must come from #Ax#.
let #Ax=x^2# #=># #A=x#
Then
#3A + B=2#
substituting #A=x#, gives
#3x + B=2#
or
#B=2-3x#
in standard from this is #B=-3x+2#.
Putting it all together we have
#x/1 + {-3x+2}/{x+3}#

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