Thread: Partial Fraction Decomposition

Looking for some help.

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

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

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

Originally Posted by RBlax

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)] ?

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

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.

Originally Posted by RBlax

Looking for some help.

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

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.)