Partial Fractions
Section 7.3
Review
Adding & Subtracting Rationals
• To decompose rational expressions into partial fractions.
4
x
18
−
−
x + 3 x − 3 x2 − 9
Review: Rational Expressions
• rational function – a quotient of two polynomials
f ( x) =
Objective
p( x)
q( x)
where p(x) and q(x) are polynomials and where q(x) is not the zero polynomial.
Definition
5y + 6
3
2y
=
+
( y − 3)( y + 2) y − 3 y 2 − y − 6
In the problem above, the fractions on the right are
called “partial fractions”. In calculus, it is often helpful
to write a rational expression as the sum of two or
more simpler rational expressions.
1
What is decomposition of partial fractions?
• Writing a more complex fraction as the sum or difference of simpler fractions.
• Examples:
3a + 4b a b
=
+
12
4 3
5x + 3 5 3
= + 2
x2
x x
• Why would you ever want to do this? It’s EXTREMELY helpful in calculus!
• And it will help you get out of MTH 112 ! ☺
4 x − 13
A(2 x − 3) + B ( x + 2)
=
( x + 2)(2 x − 3)
( x + 2)(2 x − 3)
Equate the numerators: 4x
− 13 = A(2x − 3) + B(x + 2)
Since the above equation containing A and B is true for all x, we can substitute any value of x and still have a true equation. substitute any value of x
and still have a true equation
If we choose x = 3/2, then 2x − 3 = 0 and A will be eliminated when we make the substitution. This gives us 4(3/2) − 13 = A[2(3/2) − 3] + B(3/2 + 2) −7 = 0 + (7/2)B.
B = −2.
Example 2:
Find the partial fraction decomposition.
5x − 1
( x − 3)( x + 4)
Example 1
• Decompose into partial fractions. 4 x − 13
2 x2 + x − 6
Begin by factoring the denominator:
(x + 2)(2x − 3). We know that there are constants A and B such that
4 x − 13
A
B
=
+
( x + 2)(2 x − 3) x + 2 2 x − 3
To determine A and B, we add the expressions on the right…giving us…
Example 1 continued
• If we choose x = −2, then x + 2 = 0 and B will be eliminated when we make the substitution.
• So, 4(−2) − 13 = A[2(−2) − 3] + B(−2 + 2)
−21 = −7A.
A = 3.
=3
• The decomposition is as follows:
4 x − 13
3
−2
3
2
=
+
or
−
2
2x + x − 6 x + 2 2x − 3
x + 2 2x − 3
Example 3:
Find the partial fraction decomposition.
12 x 2 + 27 x + 15
( x + 2)( x 2 − 1)
2
What if one on the denominators is a linear term squared?
Example 4: Decompose into partial fractions.
• This is accounted for by having the non‐squared term as one denominator and having the squared term as another denominator.
• What if one denominator is a linear term cubed? There would be 3 denominators in the decomposition:
would be 3 denominators in the decomposition:
7 x 2 − 29 x + 24
(2 x − 1)( x − 2) 2
A
B
C
+
+
2
(term) (term) (term)3
7 x 2 − 29 x + 24
A
B
C
=
+
+
2
(2 x − 1)( x − 2)
2 x − 1 x − 2 ( x − 2) 2
Example 4 continued
7 x − 29 x + 24 A( x − 2) + B(2 x − 1)( x − 2) + C (2 x − 1)
=
(2 x − 1)( x − 2)2
(2 x − 1)( x − 2)2
2
2
• Then, we equate the numerators. This gives us
7 x 2 − 29 x + 24 = A( x − 2)2 + B(2 x − 1)( x − 2) + C (2 x − 1)
• Since the equation containing A, B, and C is true for all of x, we can substitute any value of x and still have a true equation. In order to have 2x – 1 = 0, we let x = ½ . This gives us
1
1
1
7 ( ) 2 − 29 ⋅ + 24 = A ( − 2) 2 + 0.
2
2
2
Solving, we obtain A = 5.
Example 4 continued
In order to have x − 2 = 0, we let x = 2.
Substituting gives us 7(2) 2 − 29(2) + 24 = 0 + C (2 ⋅ 2 − 1)
C = −2
To find B, we choose any value for x except ½ or 2
and replace A with 5 and C with −2 . We let x = 1:
7⋅12 −29⋅1+ 24 = 5(1−2)2 + B(2⋅1−1)(1−2) + (−2)(2⋅1−1)
2 = 5− B − 2
B =1
Example 4 continued
Example 5:
The decomposition is as follows:
Find the partial fraction decomposition.
7 x 2 − 29 x + 24
5
1
2
=
+
−
.
2
(2 x − 1)( x − 2)
2 x − 1 x − 2 ( x − 2) 2
x+2
x( x − 1) 2
3
Example 6:
Find the partial fraction decomposition.
2x + 9
( x + 3) 2
4