Given 5/(x^2+6x+8) = A/(x+2)+B/(x+4)5x2+6x+8=Ax+2+Bx+4, what is A+BA+B ?

2 Answers
George C.
Oct 13, 2017

A+B=0A+B=0

Explanation:

Given:

5/(x^2+6x+8) = A/(x+2)+B/(x+4)5x2+6x+8=Ax+2+Bx+4

color(white)(5/(x^2+6x+8)) = (A(x+4)+B(x+2))/((x+2)(x+4))5x2+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)5x2+6x+8=A(x+4)+B(x+2)x2+6x+8

color(white)(5/(x^2+6x+8)) = ((A+B)x+(4A+2B))/(x^2+6x+8)5x2+6x+8=(A+B)x+(4A+2B)x2+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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