Okay, for example, I have 1 value which I can sub x into. And the value is 2. So now I sub x=2 and I have to find unknowns A, B and C.
After that I found A=5 and have unknown B. What should I sub x now?
Checklist:
1. The numerator has a degree less than the denominator. (Tick)
2. The denominator can be factorised. (Tick)
Now, we require a partial fraction of the form
.
The numerator is always of one less degree than the denominator.
Getting the common denominator:
.
Can you see that since the denominators are equal, so must the numerators...
So .
In this case, there does not exist an that you can use to eliminate the .
So we will use another method. Expand all the brackets...
Equating like powers of , we get
.
Can you see that for this equation to be true, then
.
Now you have three equations in three unknowns that you can solve simultaneously. Can you go from here?
Yes, if you substitute you can eliminate and to find .
But there does not exist any number that will be able to eliminate to find and .
Using your method...
.
Now that you have you should be able to solve the simultaneous equations I gave you easier.
But like I said, in this case, the ONLY method that will get you all three unknowns is to equate like powers of .
You substitute whatever will make the coefficients of A, B, C, etc = 0.
So in your case, we wanted the coefficient of to become 0.
So we let
.
Notice that the coefficient of A is . To try to eliminate A, we would need to let
.
Is it possible to square a number and end up with something negative?
You would only substitute if one of your coefficients happened to be .