Given #5/(x^2+6x+8) = A/(x+2)+B/(x+4)#, what is #A+B# ?

2 Answers
George C.
Oct 13, 2017

#A+B=0#

Explanation:

Given:

#5/(x^2+6x+8) = A/(x+2)+B/(x+4)#

#color(white)(5/(x^2+6x+8)) = (A(x+4)+B(x+2))/((x+2)(x+4))#

#color(white)(5/(x^2+6x+8)) = (A(x+4)+B(x+2))/(x^2+6x+8)#

#color(white)(5/(x^2+6x+8)) = ((A+B)x+(4A+2B))/(x^2+6x+8)#

Equating coefficients:

#{ (A+B = 0), (4A+2B=5) :}#

So:

#A+B=0#

JS
Oct 13, 2017

#A+B=0#

Explanation:

You are on the right track!

#A(x+4)+B(x+2)=5#

Now, let's set #x=-4# and solve for B:

#-2B=5#
#B=-5/2#

Now, let's set #x=-2# and solve for A:

#2A=5#
#A=5/2#

So, #A+B=0#

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