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# Partial fractions. watch

1. i have f(x) = 25/(3+2x)^2(1-x)
i need to express it as a sum of partial fractions.

So far i have done

A + B + C
(3+2x) (3+2x) (1-x)

25=A(3+2x)(1-x)+ B(3+2x)(1-x)+C(3+2x)(3+2x)
after i sub in x = 1 i get c to = 1 which is right.

However i don't get the next ones right , when i sib in x=-1.5 i get 0 ?!
2. (Original post by cuddles x)
i have f(x) = 25/(3+2x)^2(1-x)
i need to express it as a sum of partial fractions.

So far i have done

A + B + C
(3+2x) (3+2x) (1-x)

25=A(3+2x)(1-x)+ B(3+2x)(1-x)+C(3+2x)(3+2x)
after i sub in x = 1 i get c to = 1 which is right.

However i don't get the next ones right , when i sib in x=-1.5 i get 0 ?!
It supposed to be:

A + B + C
(3+2x)^2 (3+2x) (1-x)

Sub in x = 0
And x = 1 for example

You'll get two equations, with A and B as unknowns

Solve them simultaneously.
3. you dont sub in x = -1.5

you have to sub in x = -2/3, x = , 1, and then find the other 1
4. (Original post by cuddles x)
i have f(x) = 25/(3+2x)^2(1-x)
i need to express it as a sum of partial fractions.

So far i have done

A + B + C
(3+2x) (3+2x) (1-x)

25=A(3+2x)(1-x)+ B(3+2x)(1-x)+C(3+2x)(3+2x)
after i sub in x = 1 i get c to = 1 which is right.

However i don't get the next ones right , when i sib in x=-1.5 i get 0 ?!

For the sake of everyone:

5. One of the fractions needs to be (3+2x)^2 on the bottom, which gives you A(1-x) or B(1-x) on the RHS, depending on which you decide to square
6. One of the fractions should be:

A/ (3+2x)^2. You can't have two fractions with the same denominator. This will then change the line below.
7. why are one of the fractions (3+2x)^2 ?
8. (Original post by cuddles x)
why are one of the fractions (3+2x)^2 ?
Because to get to the denominator you have been given, one of the denominators had to have been (3+2x)^2

If both were (3+2x) Like you put, then the overall denominator would only be raised to the power one.
9. you should have either A or B over (3+2x)^2 and the other over (3+2x)

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