partial fractions

Partial Fractions:
Breaking up rational expressions.

Sometimes you want to take a complex rational expression and break it up into 2 or more simpler rational expressions.
This is the opposite of the way we add fractions: we find a common denominator, change both fractions to this common denominator, and add the numerators:
1/2 + 1/3 = 3/6 + 2/6 = 5/6.

For instance, what if you have
a rational expression

The denominator is actually the product of 2 binomials, (2x-7)(x+3).So we can hope that the rational expression is actually the sum of two simpler expressions. Since we don't know what the numerators are, we just use A and B for the time being.
A/(2x-7) and B/(x+3)

If we were to add these two, like we added 1/2 and 1/3 above, we would first find a common denominator (the product of the two denominators), which changes the numerators.
A(x+3)/product + B(2x-7)/product

Making this one fraction (adding numerators) and multiplying out we have a complicated expression in the numerator involving terms with x and terms without x
(Ax+3A+2Bx-7B)/(2x^2-x-21)

Combine the x and non-x terms
(A+2B)x+(3A-7B) numerator

Since the denominator is the same as the fraction we started with, the numerator must the same (though it certainly LOOKS different) and the coefficient of the x term must be the same as the 19 in the original, and the constant term must be the same as the 8 in the original.

A + 2B = 19 and
3A - 7B = 8

This is a system of equations which can be solved with either of the usual methods (substitution, here):
A = 19 - 2B
3(19 - 2B) - 7B = 8
57 - 6B -7B = 8
57 + 8 = 6B + 7B
65 = 13B
B = 5
A = 19 - 2•5 = 9

Now that we have A and B, we can substitute into the 2 partial fractions we started out with.
(It is useful to actually add these two to be sure you end up with what you started)
9/(2x-7) and 5/(x+3)

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Sometimes you want to take a complex rational expression and break it up into 2 or more simpler rational expressions. This is the opposite of the way we add fractions: we find a common denominator, change both fractions to this common denominator, and add the numerators:	1/2 + 1/3 = 3/6 + 2/6 = 5/6.
For instance, what if you have
The denominator is actually the product of 2 binomials, (2x-7)(x+3).So we can hope that the rational expression is actually the sum of two simpler expressions. Since we don't know what the numerators are, we just use A and B for the time being.
If we were to add these two, like we added 1/2 and 1/3 above, we would first find a common denominator (the product of the two denominators), which changes the numerators.
Making this one fraction (adding numerators) and multiplying out we have a complicated expression in the numerator involving terms with x and terms without x
Combine the x and non-x terms
Since the denominator is the same as the fraction we started with, the numerator must the same (though it certainly LOOKS different) and the coefficient of the x term must be the same as the 19 in the original, and the constant term must be the same as the 8 in the original.	A + 2B = 19 and 3A - 7B = 8
This is a system of equations which can be solved with either of the usual methods (substitution, here):	A = 19 - 2B 3(19 - 2B) - 7B = 8 57 - 6B -7B = 8 57 + 8 = 6B + 7B 65 = 13B B = 5 A = 19 - 2•5 = 9
Now that we have A and B, we can substitute into the 2 partial fractions we started out with. (It is useful to actually add these two to be sure you end up with what you started)