Question

Express x^3 - 2 / 2x^3 - x^2 in partial fractions.

Answer 1

Factor the denominator

Answer 2

Can you present it in a work solution?

Answer 3

Thanks.

Answer 4

This question is very tricky i cant get the 1/2 for the first term of the answer.

Answer 5

Alright, I'll try hold on.

Answer 6

Do you have the answer?

Answer 7

Answer is x^3 - 2 / x^3 + 2x = 1/2 + 4 / x + 2/x square - 15/2(2x-1)

Answer 8

@dumbcow take this >.<

Answer 9

\[\frac{x^{3}-2}{2x^{3} -x^{2}} = \frac{x^{3} -2}{x^{2}(2x-1)} = \frac{Ax+B}{x^{2}}+\frac{C}{2x-1}\]
find A,B,C

Answer 10

I tried this method already but i just couldnt get the 1/2 as the final answer as mentioned.

Answer 11

Yea, I got some weird answer with that method too.
Can't you make x, and x^2?

Answer 12

oh right sorry you need a "x^3" term
\[\rightarrow \frac{Ax^{2} +Bx+C}{x^{2}} +\frac{D}{2x-1}\]
\[(Ax^{2} +Bx+C)(2x-1) +Dx^{2} = x^{3} -2\]

Answer 13

You sure about that?

Answer 14

A = 1/2
-A +2B+D = 0 --> D = 1/2 -2B
-B +C = 0 --> C =B
C = 2

B=2
D = -7/2

Answer 15

x^\[x^3 - 2 / 2x^3 - x^2 = 1/2 + 0.5 x^2 - 2 / x^2 (2x - 1) \]

Answer 16

can you interpret this step?

Answer 17

= 1/2 + A/x + B/x^2 + C/ (2x-1)

Answer 18

I think C the answer given was 15

Answer 19

-15 sorry*

Answer 20

yes it was -15/2

i made 1 mistake

Answer 21

can you type out the whole solution if possible. Thank you

Answer 22

basically you multiply it out and then set the coefficients of each term equal to each other

Answer 23

If i were to separate the denominotor into x and x^2 would it be possible to find 1/2 as my first term answer? Thanks.

Answer 24

yes that is "A" .... A = 1/2

Answer 25

how to get the 1/2 ?

Answer 26

maybe if i write it this way
\[\frac{Ax^{2} +Bx +C}{x^{2}} = A + \frac{B}{x}+\frac{C}{x^{2}}\]
so
once you know A=1/2 , B=4, C=2, D = -15/2 from above
it becomes
\[\frac{1}{2}+\frac{4}{x}+\frac{2}{x^{2}}-\frac{15}{2(2x-1)}\]

Answer 27

great job! now i understand already :D

Answer 28

:)