Partial Fraction Decomposition
How to use the partial fraction decomposition to simplify a rational expression: formula, 3 examples, and their solutions.
Example(3x - 2)/[x(x - 1)]
The partial fraction decomposition is a way
to split a rational expression
to its partial fractions.
(= fractions with reduced denominators)
See the given expression.
The denominator is x(x - 1).
The reduced factors are
x and (x - 1).
So set
(given) = A/x + B/(x - 1).
The goal is to find A and B.
Change the denominators
to x(x - 1).
Add and Subtract Rational Expressions
See A/x.
The denominator is x.
The factor (x - 1) is missing.
So multiply (x - 1)
to both of the numerator and the denominator.
A/x
= [A/x]⋅[(x - 1)/(x - 1)]
Next, see +B/(x - 1).
The denominator is (x - 1).
The factor x is missing.
So multiply x
to both of the numerator and the denominator.
+B/(x - 1)
= [B/(x - 1)]⋅[x/x]
So
A/x + B/(x - 1)
= [A/x]⋅[(x - 1)/(x - 1)] + [B/(x - 1)]⋅[x/x].
A(x - 1) = Ax - A
Multiply a Monomial and a Polynomial
+B⋅x = +Bx
Ax + Bx = (A + B)x
Common Monomial Factor
(3x - 2)/[x(x - 1)] = [(A + B)x - A]/[x(x - 1)]
The denominators are the same.
So the numerators are also the same.
The x terms are the same.
So the coefficients are the same.
So A + B = 3.
The constant terms are the same.
So -A = -2.
A + B = 3
-A = -2
Find A and B
by solving this system.
-A = -2
So A = 2.
Put A = 2
into A + B = 3.
Then 2 + B = 3.
Substitution Method
Move 2 to the right side.
Then B = 1.
(given) = A/x + B/(x - 1)
A = 2
B = 1
So (given) = 2/x + 1/(x - 1).
So
2/x + 1/(x - 1)
is the answer.
Example(5x2 - 1)/[x2(x - 1)]
See the given expression.
The denominator is x2(x + 1).
The reduced factors are
x, x2, and (x + 1).
So set
(given) = A/x + B/x2 + C/(x + 1).
The goal is to find A, B, and C.
Change the denominators
to x2(x + 1).
See A/x.
The denominator is x.
x(x + 1) is missing.
So multiply x(x + 1)
to both of the numerator and the denominator.
A/x
= [A/x]⋅[[x(x + 1)]/[x(x + 1)]]
See +B/x2.
The denominator is x2.
(x + 1) is missing.
So multiply (x + 1)
to both of the numerator and the denominator.
+B/x2
= +[B/x2]⋅[(x + 1)/(x + 1)]
See +C/(x + 1).
The denominator is (x + 1).
x2 is missing.
So multiply x2
to both of the numerator and the denominator.
+C/(x + 1)
= +[C/(x + 1)]⋅[x2/x2]
So
A/x + B/x2 + C/(x + 1)
= [A/x]⋅[[x(x + 1)]/[x(x + 1)]] + [B/x2]⋅[(x + 1)/(x + 1)] + [C/(x + 1)]⋅[x2/x2].
A⋅x(x + 1)
= Ax(x + 1)
= Ax2 + Ax
+B(x + 1) = +Bx + B
+C⋅x2 = +Cx2
Ax2 + Cx2 = (A + C)x2
Ax + Bx = (A + B)x
(5x2 - 1)/[x2(x + 1)]
= [(A + C)x2 + (A + B)x + B]/[x(x + 1)]
The denominators are the same.
So the numerators are also the same.
The x2 terms are the same.
So the coefficients are the same.
So A + C = 5.
There's no x term in the given expression.
So (A + B)x = 0x.
So A + B = 0.
The constant terms are the same.
So B = -1.
A + C = 5
A + B = 0
B = -1
Find A, B, and C
by solving this system.
System of Equations (3 Variables)
B = -1
One variable is already known.
Put B = -1
into A + B = 0.
Then A + (-1) = 0.
+(-1) = -1
Move -1 to the right side.
Then A = 1.
Put A = 1
into A + C = 5.
Then 1 + C = 5.
Move 1 to the right side.
Then C = 4.
(given) = A/x + B/x2 + C/(x + 1)
A = 1
B = -1
C = 4
So (given) = 1/x + (-1)/x2 + 4/(x + 1).
+(-1)/x2 = -1/x2
So
1/x - 1/x2 + 4/(x + 1)
is the answer.
FormulaSimplify 1/AB
1/AB = [1/(B - A)]⋅[1/A - 1/B]
Use this formula
when (B - A) doesn't have a variable.
(= constant)
Example1/[x(x + 1)]
The denominator is x(x + 1).
(x + 1) - x = 1
It doesn't have a variable.
So use the partial fraction decomposition formula.
1/[x(x + 1)] = [1/[(x + 1) - x]]⋅[1/x - 1/(x + 1)]
(x + 1) - x = 1
[1/1]⋅[1/x - 1/(x + 1)]
= 1/x - 1/(x + 1)
So
1/x - 1/(x + 1)
is the answer.