# Partial Fraction Decomposition. [closed]

The intial fraction is:

$\dfrac{3x^3-x+4}{x^3+2x^2+6x}$

How should to make the decomposition for the denominator?

Thanks!

First divide the numerator by the denominator!

$$\frac{3x^3-x+4}{x^3+2x^2+6x}=3+\frac{-6x^2-19x+4}{x^3+2x^2+6x}$$

Then note that the denominator factors, at best, to: $$x^3 + 2x^2 + 6x = x(x^2 + 2x + 3)$$

$$3+\frac{-6x^2-19x+4}{x^3+2x^2+6x} = 3 + \dfrac{A}{x} + \dfrac{Bx + C}{(x^2 + 2x + 3)}$$

• Make sense! Thanks! Dec 9 '13 at 18:38
• @amWhy: Nice work +1 (I see that your stalker is doing wrong by DV correct answers - petty and low)! Dec 10 '13 at 3:02

Note that since the numerator and denominator of the fraction have the same order so we can divide them to find out:$$\frac{3x^3-x+4}{x^3+2x^2+6x}=3+\frac{-6x^2-19x+4}{x^3+2x^2+6x}$$ Now follow @amWhy's hint.

$\dfrac{3x^3-x+4}{x^3 + 2x^2 + 6x} = 3+\dfrac{-6x^2-19x+4}{x^3+2x^2+6x}$ $$\\$$ $\dfrac{-6x^2-19x+4}{x^3+2x^2+6x}$ = $\dfrac{-6x^2-19x+4}{x(x^2+2x+6)}$ = $\dfrac{A}{x} + \dfrac{Bx+C}{x^2+2x+6}$

You can get the value of A directly by putting $x = 0$ in this polynomial : $\dfrac{-6x^2-19x+4}{x^2+2x+6}$ . That is the value of A is $\dfrac{4}{6}$

$$\frac{(3x^3-x+4)}{x^3 + 2x^2 + 6x}$$

First you have to notice that the highest power in the numerator is equal to the highest power in the denominator. So you have to do synthetic division to take out a factor from the numerator:

$3x^3+0x^2-x+4 \overline{)x^3 + 2x^2 + 6x+0}$

You'll end up with: $factor + \frac{(...)}{x(x^2 + 2x + 6)}$ where (...) is your remainder after long dividing.

Then you need to factor the denominator.

$x^3 + 2x^2 + 6x = x(x^2+2x+6)$

$x^2+2x+6$ is irreducible so your partial fraction decomposition will look like:

$$\frac{(...)}{x(x^2 + 2x + 6)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 2x + 6}$$

You'll set the corresponding coefficients on the left and right equal to each other and solve for A,B,C,D (usually using a matrix).