# Thread: Partial fractions, long division.

1. ## Partial fractions, long division.

so the question has two parts and goes like this:

1) use long division to turn the rational function:

f(x)=(2x^4-5x^3-9x-12)/(x^3-3x^2+3x-9)

into the sum of a polynomial and a proper rational function g(x).

so i did this and got the answer, then i got stuck on part two when it asked:

2) Look for a small integer root of the denominator of f(x); and use this to find the partial fraction decomposition of g(x)

2. Hi Sniperpro

The question asks you to find the root of (x^3-3x^2+3x-9), then you factorize the denominator. Finally, find the partial fraction decomposition of g(x)

3. Originally Posted by Sniperpro
so the question has two parts and goes like this:

1) use long division to turn the rational function:

f(x)=(2x^4-5x^3-9x-12)/(x^3-3x^2+3x-9)

into the sum of a polynomial and a proper rational function g(x).

so i did this and got the answer, then i got stuck on part two when it asked:

2) Look for a small integer root of the denominator of f(x); and use this to find the partial fraction decomposition of g(x)

4. 2x+1-[(3x^2+6x-3)/(x^3-3x^2+3x-9)]

5. i found the root to be x=3, so does this mean i can substitute x^3-3x^2+3x-9 with (x-3)^3?

6. Hi sniperpro
Originally Posted by Sniperpro
i found the root to be x=3, so does this mean i can substitute x^3-3x^2+3x-9 with (x-3)^3?
No, you can't because (x-3)^3 $\displaystyle \neq$ x^3-3x^2+3x-9. Because you know that x = 3 is the root, you can use Horner's method or long division to factorize x^3-3x^2+3x-9

7. Originally Posted by Sniperpro
2x+1-[(3x^2+6x-3)/(x^3-3x^2+3x-9)]
should be ...

$\displaystyle 2x+1 - \frac{3x^2 - 6x + 3}{x^3 - 3x^2 + 3x - 9}$

note that the denominator of the original fraction (and the last term above) equals 0 when x = 3 ... according to the factor theorem, that makes (x-3) a factor ...

$\displaystyle 2x+1 - \frac{3x^2 - 6x + 3}{(x-3)(x^2+3)}$

the last question also wants a partial fraction decomposition of the last term ...

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

your last task is to find the values of the constants A , B, and C.

8. long division to factorize?

9. Originally Posted by Sniperpro
long division to factorize?
factorization is just an intermediate process toward the ultimate goal of the partial fraction decomposition.

10. Hi Sniperpro

As skeeter said :
note that the denominator of the original fraction (and the last term above) equals 0 when x = 3 ... according to the factor theorem, that makes (x-3) a factor ...
So, $\displaystyle \frac{x^3 - 3x^2 + 3x - 9}{x-3}=x^2 + 3$ and you can factorize x^3 - 3x^2 + 3x - 9.

x^3 - 3x^2 + 3x - 9 = (x-3)(x^2 + 3) as skeeter has pointed out ^^