# `10/(5x^2 - 2x^3)` Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients.

*print*Print*list*Cite

### 2 Answers

`10/(5x^2-2x^3)`=`10/(x^2(5-2x))` ``

=`A/x+B/x^2+C/(5-2x)`

= `(Ax(5-2x)+B(5-2x)+Cx^2)/(5x^2-2x^3)`

`` `` `10=Ax(5-2x)+B(5-2x)+Cx^2`

`10=5Ax-2Ax^2+5B-2Bx+Cx^2`

Now we have to equate coefficients of same power terms on both sides we get,

Equating coefficients of `x^2` we get,

-2A+C=0

i.e C=2A ------> (1) ``

Equating coefficients of x we get,

5A-2B=0 ------->(2)

Equating constants we get,

5B=10

i.e B=2

Substituting value of B in (2) we get,

5A-4=0

i.e `A=4/5`

Now substituting value of A in (1) we get,

C=2( 4/5)=8/5

Therefore,

`10/(5x^2-2x^3)=4/(5x)+2/(x^2)+8/(5(5-2x))`

This is the partial fraction decomposition.

### User Comments

`10/(5x^2 - 2x^3)`

**sol:**

Given

`10/(5x^2 - 2x^3)`

to find the partial decomposition we need to consider the denominator first and factorize it

so,

`(5x^2 - 2x^3) = ((x^2)*(5-2x))`

so the partial decomposition of `10/(5x^2 - 2x^3)` is

`10/(5x^2 - 2x^3) = A/(5-2x) + B/(x) + C/(x^2)`

:)