Decompose a fraction in a sum of two Let's say that I have this fraction:
$$ \frac{2x}{x^2+4x+3}$$
I would like to decompose in two fraction:
$$ \frac{A}{x+3} + \frac{B}{x+1}$$
Which is the procedure for that?  :)
 A: Multiply either sides by $x^2+4x+3=(x+1)(x+3)$ and compare the constants and the coefficients of $x$ of the numerator to form two linear simultaneous equation for unknown $A,B$
See  Partial Fraction Decomposition
A: $$ \frac{2x}{x^2+4x+3}=\frac{2x}{x^2+x+3x+3}=\frac{2x}{x(x+1)+3(x+1)}=$$
$$=\frac{2x}{(x+3)(x+1)}=\frac{A}{x+3}+\frac{B}{x+1}$$
$$2x=Ax+A+Bx+3B=(A+B)x+A+3B$$
You needs to solve the system
$$A+B=2,A+3B=0$$
A: Add both:
$$\frac{2x}{x^2+4x+3}=\frac{Ax+A+Bx+3B}{x^2+4x+3}$$
$$\frac{2x}{x^2+4x+3}=\frac{(A+B)x+(A+3B)}{x^2+4x+3}$$
Then how this must be valid for all $x$ then $A+B=2$ and $A+3B=0$
A: Hint:
$$\frac{A}{x+3} + \frac{B}{x+1} = \frac{A(x+1) + B(x+3)}{(x+1)(x+3)} = \frac{(A+B)x + (A+3B)}{x^2+4x+3} = \frac{2x}{x^2+4x+3}$$
What must the values of $A$ and $B$ so that $A+B = 2$ and $A + 3B = 0$?
A: $$x^2+4x+3=0$$
$$\Delta=4^2-4 \cdot 1 \cdot 3=4$$
$$x_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-4 \pm 2}{2}=-3 \text{ or } -1$$
$$x^2+4x+3=(x+3)(x+1)$$
$$\frac{2x}{x^2+4x+3}=\frac{2x}{(x+3)(x+1)}=\frac{A}{x+3}+\frac{B}{x+1}=\frac{A(x+1)+B(x+3)}{(x+3)(x+1)}=\frac{(A+B)x+(A+3B)}{(x+3)(x+1)}$$
Since,$\frac{2x}{(x+3)(x+1)}=\frac{(A+B)x+(A+3B)}{(x+3)(x+1)}$,it must be:
$$A+B=2 \text{ and } A+3B=0$$
$$A+3B=0 \Rightarrow A=-3B$$
$$A+B=2 \Rightarrow -2B=2 \Rightarrow B=-1$$
$$A=3$$
Therefore:
$$\frac{2x}{x^2+4x+3}=\frac{3}{x+3}-\frac{1}{x+1}$$
