# Stuck on decomposing partial fraction

I want to decompose the following, and I think got stuck in the thick of it $$\dfrac{2x^3+3x+1}{(x+1)^2}$$

I tried like this:

OK, after advice from @Daniel Fischer and @lab bhattacharjee I decided to use division:

first separated the equation as $[\dfrac{2x^3+3x+1}{(x+1)}]\times \dfrac{1}{(x+1)}$

The I used polynomial division on $[\dfrac{2x^3+3x+1}{(x+1)}]$ to get $[2x^2-2x+3- \dfrac{2}{x+1}] \times \dfrac{1}{(x+1)}$

with me, $[2x^2-2x+3- \dfrac{2}{x+1}] \times \dfrac{1}{(x+1)}$ reduces to $\dfrac{2x^2-2x+3}{x-1}-2$

Is this still on course, so far?

• For a partial fraction decomposition, the degree of the numerator must be smaller than the degree of the denominator. So first you must do a polynomial division. – Daniel Fischer Oct 4 '13 at 9:10

## 2 Answers

Your polynomial long division is wrong: instead of factoring out a $1/(x+1)$, just divide $$(2x^3 + 3x + 1)\div (x^2 + 2x + 1) = 2x - 4 + \frac{r(x)}{(x+1)^2}.$$ Now use partial fractions on the remainder term: $$r(x) = A(x+1) + B = (A)x + (A+B)$$ and equate the coefficients on both sides to get a system in $A$ and $B$. Solve the system.

Answer: $r(x) = 9x + 5$, $A = 9$, $B = -4$

As the highest power of numerator is greater that that of denominator

$$\dfrac{2x^3+3x+1}{(x+1)^2}=2x+A+\frac B{x+1}+\frac C{(x+1)^2}$$

On multiplication by $(x+1)^2$ we get $$2x^3+3x+1=(x+1)^2(2x+A)+B(x+1)+C$$

$$\implies 2x^3+3x+1=2x^3+x^2(4+A)+x(2+2A+B)+A+B+C$$

Now compare the coefficients of the different powers of $x$ to find $A,B,C$

• @Sylvester, please find the edited version – lab bhattacharjee Oct 4 '13 at 18:04