Partial Fractions
Written by Don Methven and generously donated to the Math and Physics Help Home Page
Partial fractions: I personally find these to be quite tough - especially if you aren't good with fractions in the first place. They
involve the splitting up of a fraction into two or more fractions with only one factor in the
denominator. I have written this as a quick reminder on how you use the methods to cope with them. Basically I have solved
a partial fraction with each method, so you can try it yourself. I would advise you to print this
out and follow it along (maths is hard to do on a screen).
OK lets go...
Writing:
6 1 1
------------- = ------- - -------
x² + 2x - 8 x - 2 x + 4
6
means that you have expressed ----------- in partial fractions...
x² + 2x - 8
Now, there are three main "methods" of doing this:
(1) Substitution of strategic values
(2) Solving with co-efficiants
(3) Cover up method (can only be used on fractions without powers in the bottom bit)
Method 1
Substitution of values.
=======================
Express as a partial fraction.
x - 1
---------------
(3x - 5)(x - 3)
=======================
First write the fraction as:
x - 1 A B
--------------- = -------- + -------
(3x - 5)(x - 3) (3x - 5) (x - 3)
Notice that I have taken the two terms that are in brackets and placed them on there own,
in there own fraction (the A and B are what we need to find out)
next, multiply denominator (the bottom bit of the big fraction) by both sides:
so we now have:
x - 1 = A(x - 3) + B(3x - 5)
to get rid of one term, substitute a 'strategic' value - example: to get rid of (x - 3), make x = 3, so (3 - 3) ends up as 0 - its gone! Remember that you have to do this to all the x's in the equation though.
3 - 1 = B(3*3 - 5)
2 = 4B
so, 2/4 = B
or, 1/2 = B
now, do the same except this time get rid of B leaving A behind.
- make x = 5/3
so, (5/3 * 3) = 5
so, (5 - 5) = 0
- the (3x - 5) term is gone!
5/3 -1 = A(5/3 - 3)
2/3 = A * -4/3
2/3
------ = A
-4/3
-1/2 = A
so, we now have A and B - the answer!
x - 1 -1 1
--------------- = --------- + --------
(3x - 5)(x - 3) 2(3x - 5) 2(x - 3)
#########################################################################################
Method 2
Solving by co-efficiants
========================
Usually people use this method in conjunction with another method ie substitution of values, there is however nothing wrong with using this method on its own if you prefer it - as below...
=======================
Express as a partial fraction.
x - 1
---------------
(3x - 5)(x - 3)
=======================
First write the fraction as:
x - 1 A B
--------------- = -------- + -------
(3x - 5)(x - 3) (3x - 5) (x - 3)
The first step (as in the first method) is to multiply the denominator by both sides
x - 1 = A(x - 3) + B(3x - 5)
It may help if the A and B parts are expanded but this step can usually be missed.
x - 1 = A * x - A * 3 + B * 3 * x - 5B
x - 1 = Ax - 3A + 3Bx - 5B
now look at this and try to equate co-efficiants for 'x'
Ok, we've now done the first step, now do the same agian but with somthing else (like x² or
constants (the constants are the actual number's - i.e not x's which can be anything)
Lets do, co-efficiant for constants.
-1 = -3A -5B
we now have a simultanious equation...
1 = A + 3B --- (1)
-1 = -3A + (-3B) --- (2)
Solve.
multiply (1) by -3
-3 = -3A + (-9B) --- (1a)
subtract (1a) from (2)
-2 = -4B
-2
------- = B
-4
0.5 = B
plug into (1)
1 = A + 3 * 0.5
1 - 3 * 0.5 = A
-0.5 = A
we now know A and B
x - 1 0.5 -0.5
--------------- = --------- + --------
(3x - 5)(x - 3) (3x - 5) (x - 3)
this can be written with "1/2's" instead of the "0.5's"
x - 1 -1 1
--------------- = --------- + --------
(3x - 5)(x - 3) 2(3x - 5) 2(x - 3)
#########################################################################################
Method 3
cover up method
================
This method is in my opinion the easiest of all the three methods, but it can be misleading if you follow it like a cookbook recipe, not knowing what you are really doing. It also cannot be used with non-linear fractions such as explained after this section...I basically just done a partial fraction not explaing it much - since you can pretty much see how you do it from example.
=======================
Express as a partial fraction.
x - 1
---------------
(3x - 5)(x - 3)
=======================
delete first term.
3x - 5
3x - 5 = 0
3x = 5
x = 5/3
plug 5/3 as x into what is left.
5/3 - 1 2/3 -1
------------- = ------------ = ---
5/3 - 3 -4/3 2
this is the answer for the (3x - 5) expression
repeat for other expression.
delete second term.
x - 3
x - 3 = 0
x = 3
plug in.
3 - 1 2 1
----------- = --- = ---
3*3 - 5 4 2
this is the answer for the (x - 3) expression.
so,
x - 1 -1 1
--------------- = --------- + --------
(3x - 5)(x - 3) 2(3x - 5) 2(x - 3)
done.
pretty quick huh?
#########################################################################################
Strange exceptions (repeated-linear fractions)
for a repeated linear fraction, i.e. (1 + x)2 the format will look somthing like:
A B C
fraction = ------------ + --------- + ---------
expression (1 + x) (1 + x)²
notice that becuase (1 + x)2 is in the 'expression' it also has an (1 + x) not squared to go with it
(obviously there will not be a 1 + x in every expression and the number is merely a representation).
example:
=======================
Express as a partial fraction.
1
---------------
(x - 3)(x + 1)²
=======================
1 A B C
--------------- = ------- + ------- + --------
(x - 3)(x + 1)² (x - 3) (x + 1) (x + 1)²
multiply by the denominator.
1 = A(x + 1)² + B(x - 3)(x + 1) + C(x - 3)
multiply out (in this case its only the 'A' term).
1 = A(x +1)(x +1) + B(x - 3)(x + 1) + C(x - 3)
solve by making x = -1 (this is an 'inspection method' part)
1 = C(-1 - 3)
1 = C*4
-1/4 = C
make x = 3
1 = A(3 + 1)(3 + 1)
1 = A*16
1/16 = A
solve coefficients of x2 for B
0 = A + B
0 = 1/16 + B
-1/16 = B
solved.
1 1 1 1
--------------- = ----------- - ----------- - ------------
(x - 3)(x + 1)² 16(x - 3) 16(x + 1) 4(x + 1)²
#########################################################################################
Strange exceptions (quadratics in the denominator - bottom of the fraction - )
If you have got a fraction in the denominator i.e. x² + 3x + 2 the format should look like this:
A Bx + C
fraction = --------- + ------------
factor quadractic
example:
=======================
Express as a partial fraction.
x - 1
--------------------
(x + 3)(x² + 3x + 2)
=======================
x - 1 A Bx + C
-------------------- = --------- + -------------
(x + 3)(x² + 3x + 2) x + 3 x² + 3x + 2
multiply by denominator.
x - 1 = A(x² + 3x + 2) + (Bx + C)(X + 3)
By Inspection
make x = -3 (at this point be careful that your selected value doesn't also make the
quadratic equal 0 as well!)
-3 - 1 = A((-3)² + 3*(-3) + 2)
-4 = A(9 - 9 + 2)
-4 = 2A
-2 = A
solve co-efficiant of x²
0 = A + B
0 = -2 + B
2 = B
solve co-efficiants of x
1 = 3A + 3B + C
1 = 3*(-2) + 3*(2) + C
1 = C
done.
x - 1 -2 2x + 1
-------------------- = --------- + ---------------
(x + 3)(x² + 3x + 2) (x + 3) x² + 3x + 2
Hope all of that helped!
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