Now, you can ignore the denominators and multiply out the numerator on the right and compare terms with the left to find A, B and C.
The reason recommended now finding f(0), f(1) and f(-0.5) is because f(0) finds A since the parts with B and C will be zero.
You get and so on for B and C.
For more examples of partial fractions:
Partial fraction - Wikipedia, the free encyclopedia
Helolo, HNCMATHS!
I'll show you the method I was taught.
I think it's much simpler; don't know why everyone doesn't teach it.
Set up the equation: .Resolve into partial fractions: .
Multiply by the LCD: .
Select some "good" values for
Let . . . . . . . . . . . . .
Let . . . . . . . . . . . .
Let . . .
Therefore: .
[Edit: oops! .I see that recommended this method.]
My main problem is that i dont understand how to work out that values that we need to plug in for
EDIT**
I think I have worked it out, has to equal 0 in the divisor? so in our case where we have ..... would equal
Its just clicked, again thank you all and i will be clicking thanks on all your posts.