Question

Partial fractions help.

Answer 1

.......yes??

Answer 2

will someone please set up (4 5 or 6) and do one of 7 and 8 please?

Answer 3

lets do 4 :)

Answer 4

3
-----------
(2x+1)(3x-2)

Answer 5

We need on "fraction" for each factor in the bottom.

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

is thsi correct?

Answer 6

That looks good

Answer 7

After we split it up into its "baser" units, we can try to "re-solve" it.

Answer 8

A(3x-2) B(2x+1)
----------- + ----------- correct?
(2x+1)(3x-2) (3x-2)(2x+1)

Answer 9

the bottoms become redundant and we want the tops to equal out.

Answer 10

Are you asking me whether you're right or are you just putting the questions there?

Answer 11

guys no 4 is just setting up

Answer 12

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

lol.... they are really rhetorical, but I do tend to mess up alot ;)

Answer 13

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

combine like terms...

Answer 14

3x(A) + 2x(B) -2(A) +1(B) ; factor out the variables now

Answer 15

x(3A +2B) -1(2A +B) = 0x + 3

Answer 16

3A +2B = 0

-2A -B = 3

Answer 17

we can solve for A and B thru a system of equations now

Answer 18

B = -3-2A

3A +2(-3-2A) = 0

3A -6 -4A = 0

-A = +6

A = -6

Answer 19

B = -3-2(-6)

B = -3+12

B = 9

Answer 20

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



-3 9 3
------ + ------ = -----------
(2x+1) (3x-2) (2x+1)(3x-2)

Answer 21

and thats our decomposition; any questions?

Answer 22

for the set up only where should i stop?

Answer 23

you should stop when all the possible "denominators" are accounted for: for example,

if the denominator is (x+2)^3 we need to account for..

(x+2) and (x+2)^2 and (x+2)^3 as possible denominators in our setup

Answer 24

what do we do for 3 terms in numerator?

Answer 25

the 3 in the numberator is what we use to set up the condition for our "unknowns".

A + B have to end up equaling 3 in the end.... its our goal, our guide, our calibration.

Answer 26

If we get stuff like x^2(A+B) and there is no value for x^2 in our numberator, we set it up by 0x^2

Answer 27

I might be missing a "+C" in there but I cant remember :)

Answer 28

can you do 5 or 6 please? i am confused

Answer 29

I can...... :)

Answer 30

3x^2 +x -1
--------------
(x^2+2) (x^2+1)

can we factor the bottom any more? I dont see that we can....

Answer 31

no

Answer 32

A B 3x^2 +x -1
------- + ------- = ----------
(x^2+2) (x^2+1) yadayada

Try to "solve" by getting like denominators now :) like normal...

Answer 33

A(x^2+1) +B(x^2+2) = 3x^2 +x -1
------------------ -----------
yadayada yadayada

makes sense?

Answer 34

yeah. thank you. I just needed the concept.

Answer 35

now we can "turn" that left side into the right side like this :)

Answer 36

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

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

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

A + B = 3
-------------------

Answer 37

A + 2B = -1

A = -1 -2B
------------------

Answer 38

-1 -2B +B = 3

-B = 4

B = -4

Answer 39

i get it now.

Answer 40

thank you very much

Answer 41

gtg now

Answer 42

seeya

Answer 43

youre welcome :) bye..