N(x) rest(x)
---- = Quotient(x) + --------
D(x) D(x)
Example
3x2 + 4x 7
---------- = (3x + 7) + ---------
x - 1 x - 1
A A
------- ; --------- ;
(x - a) (x - a)n
Ax + B Ax + B
-------------- ; ----------------- with b2 - 4c < 0
x2 + bx + c (x2 + bx + c)n
Here A, B, b and c are real constants.
Examples
5
---------
(x -2)2
2x + 3
------------
x2 + x+ 4
2x + 3
----------------
(x2 + x+ 4)2
(x - a)n and/or (x2 + bx + c)n
The proper rational fraction can be written as a sum of partial fractions. The nature and the number of partial fractions depends on the nature of the factors in the denominator.Each factor (x - a)n in the denominator causes a sum of n partial fractions A B C L ------- + ------- + ------- + ... + ------- 2 3 n (x - a) (x - a) (x - a) (x - a) Each factor (x2 + bx + c)n in the denominator causes the sum of n partial fractions Ax + B Cx + D Lx + M --------------- + --------------- + ...+ --------------- 2 2 2 2 n (x + bx + c) (x + bx + c) (x + bx + c)
2x + 5
--------------------
x3 - 3x2 - 4x + 12
We factor the denominator x3 - 3x2 - 4x + 12 = (x-3)(x+2)(x-2)
Each factor causes exactly 1 elementary fraction of the sum
2x + 5 A B C
---------------- = ------- + ------- + -------
(x-3)(x+2)(x-2) (x - 3) (x + 2) (x - 2)
A, B and C are presently unknown. Now we'll show how to calculate A, B and C.
First we write the right side with one common denominator.
Then the denominators are equal on both sides.
So, the numerators must be equal.
(2x + 5) = A(x + 2)(x - 2) + B(x - 3)(x - 2) + C(x - 3)(x + 2)
(2x + 5) = (A + B + C)x.x + (-5B - C)x + (-4A +6B -6C)
<=>
/ A + B + C = 0
| -5B - C = 2
\ -4A +6B -6C=5
<=>
. . .
<=>
A = 2.2 B = 0.05 C = -2.25
Conclusion:
2x + 5 2.2 0.05 2.25
---------------- = ------- + ------- - -------
(x-3)(x+2)(x-2) (x - 3) (x + 2) (x - 2)
7x
---------------------
x3 - 6x2 + 12x - 8
We factor the denominator and we find (x - 2)3
This factor causes a sum of three partial fractions
7x A B C
-------- = ------- + ------- + -------
3 2 3
(x - 2) (x - 2) (x - 2) (x - 2)
A, B and C are presently unknown. Now we'll show how to calculate A, B and C.
First we write the right side with one common denominator.
Then the denominators are equal on both sides.
So, the numerators must be equal.
7x = A (x - 2)2 + B (x-2) + C
<=>
7x = A x2 + (B - 4A) x + 4A - 2B + C
with the same method as in previous example we calculate A, B and C.
A = 0 ; B = 7; C = 14
Conclusion:
7x 7 14
-------- = ------- + -------
3 2 3
(x - 2) (x - 2) (x - 2)
2 2
5x + 4x + 1 5x + 4x + 1
----------------- = -----------------
3 2 2 2 3
(x + x )(x + 1) (x )(x + 1)
The factor x2 in the denominator causes a sum of two partial fractions.
The factor (x + 1)3 in the denominator causes a sum of three partial fractions.
A B C D E
--- + ---- + ------- + ------- + -------
2 2 3
x x (x + 1) (x + 1) (x + 1)
A, B, C, D and E are presently unknown. Now we'll show how to calculate A, B, C, D and E
First we write the right side with one common denominator.
Then the denominators are equal on both sides.
So, the numerators must be equal.
5 x2 + 4x + 1 = A x (x + 1)3 + B (x + 1)3 + C x2 (x + 1)2 + D x2 (x + 1) + E x2
<=>
5 x2 + 4x + 1 =(A + C) x4 + (3 A + B + 2C + D)x3 + (3A + 3B + C + D + E)x2 + (A + 3B)x + B
<=>
/ A + C = 0
| 3 A + B + 2C + D = 0
| 3 A + 3B + C + D + E = 5
| A + 3B = 4
\ B = 1
<=>
...
<=>
A = 1 ; B = 1; C = -1 ; D = -2 ; E = 2
Conclusion:
1 1 -1 -2 2
--- + ---- + ------- + ------- + -------
2 2 3
x x (x + 1) (x + 1) (x + 1)
(2x + 1)
------------------------
2 2
( x + 1)( x + x + 1)
The denominator can't be factored in real factors of lower degree.
Each factor in the denominator gives us just one elementary fraction with a numerator
of the first degree.
(2x + 1) Ax + B Cx + D
---------------------- = --------------- + -----------------
2 2 2 2
( x + 1)( x + x + 1) ( x + 1) ( x + x + 1)
A, B, C and D are presently unknown. Now we'll show how to calculate A, B, C, and D.
First we write the right side with one common denominator.
Then the denominators are equal on both sides.
So, the numerators must be equal.
(2x + 1) = (Ax + B)(x2 + x + 1) + (Cx + D)(x2 + 1)
<=>
(2x + 1) = (A + C) x3 + (A + B + D) x2 + (A + B + C) x + B + D
and with the aid of a system with 4 unknowns, we find
...
A = -1 ; B = 2 ; C = 1; D = -1
Conclusion
(2x + 1) - x + 2 x - 1
---------------------- = --------------- + -----------------
2 2 2 2
( x + 1)( x + x + 1) ( x + 1) ( x + x + 1)
x4 + 5x3 + 16x2 + 26x + 22
------------------------------
x3 + 3 x2 + 7x + 5
This improper fraction can be written as the sum of a polynomial and a
proper rational fraction.
3 x2 + 7 x + 12
(x+2) + -------------------------
x3 + 3 x2 + 7x + 5
We factor the denominator in (x+1)(x2 + 2x + 5)
3 x2 + 7 x + 12 A B x + C
---------------------- = ------------ + -----------------
(x+1)(x2 + 2x + 5) x + 1 x2 + 2x + 5
A, B and C are presently unknown.
First we write the right side with one common denominator.
Then the denominators are equal on both sides.
So, the numerators must be equal.
3x2 + 7x + 12 = A(x2 + 2x + 5) + (Bx + C)(x + 1)
As above, you can find A, B and C by solving a system of three equations.
A = 2 ; B = 1 ; C = 2
3 x2 + 7 x + 12 2 x + 2
---------------------- = ------------ + -----------------
(x+1)(x2 + 2x + 5) x + 1 x2 + 2x + 5
16 1 1 4
------------------ = ----------- - ----------- - ------------
x3 - x2 - 5x - 3 (x - 3) (x + 1) (x + 1)2
- 3 x3 + 8 x2 - 4 x + 5 x 3 1
---------------------------------- = ------------ + ------- - ------
- x4 + 3 x3 - 3 x2 + 3 x - 2 x2 + 1 x - 1 x - 2
2 x3 + 7 x2 - 2 x + 6 2x 3
------------------------ = ------------------ + ---------------
x4 + 4 x2 - 2 x + 2 x2 + 2 x + 2