Given #5/(x^2+6x+8) = A/(x+2)+B/(x+4)#, what is #A+B# ?
2 Answers
Oct 13, 2017
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#
Oct 13, 2017
Explanation:
You are on the right track!
Now, let's set
Now, let's set
So,