1. ## Partial Fraction Decomposition

$\displaystyle (8x+12)/x(x+4) =A/x+B/(x+4)$

Looking for some help.

This is how I'm approaching it so far and not sure I'm doing it correctly.

$\displaystyle 8x+12=A(x+4)+B(x)$

$\displaystyle 8x+12=(A+B)x+(4A+B)$

2. You are proceeding correctly. Your next step would be to note that the coefficients of powers of x must be equivalent on each side of the equation, as A and B are numbers, not functions of x, and cannot introduce an x term.
That is, you are now to solve the two simultaneous equations:
8 = A + B
12 = 4A + B

3. So then x=1, A = 4/3, B = 20/3 ??

4. Originally Posted by RBlax
$\displaystyle (8x+12)/x(x+4) =A/x+B/(x+4)$
Is the left side correctly stated: (8x+12)/x(x+4) means [(8x+12)/x] * (x+4)
or do you mean (8x+12) / [x(x+4)] ?

5. (8x+12) over (x(x+4)) =)

6. My apologies; I did not actually check your algebra in your first post.
In the line:
8x + 12 = A(x + 4) + Bx
the next line should be:
8x + 12 = (A + B)x + 4A
so that the simultaneous equations are:
A + B = 8
4A = 12.
Note that you need not rely on us to check whether your solution is correct; simply add the resulting fractions together replacing A and B as necessary and see if you get the correct result.

7. Originally Posted by RBlax
$\displaystyle (8x+12)/x(x+4) =A/x+B/(x+4)$

Looking for some help.

This is how I'm approaching it so far and not sure I'm doing it correctly.

$\displaystyle 8x+12=A(x+4)+B(x)$

$\displaystyle 8x+12=(A+B)x+(4A+B)$
As you have been told, two polynomials are the same for all x if and only if they have the same coefficients: you must have A+ B= 8 and 4A+ B= 12.

Here's a slightly different way: going back to 8x+ 12= A(x+ 4)+ Bx, since this is true for all x, you can choose whatever values you like for x. In particular, taking x= -4 and then x= 0 gives you two very simple equations!

(You can't always find values of x that will make it that easy- in that case, you set coefficients equal or just choose two simple values for x to get two equations to solve for A and B.)