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Partial Fractions - Cover Up Rule | Brilliant Math & Science Wiki
Partial Fractions - Cover Up Rule
In partial fraction decomposition, the cover-up rule is a technique to find the coe cients of linear terms in a partial fract
decomposition. It is a faster technique in finding constants in a partial fraction. We can only apply this rule when the
denominator is a product of linear factors.
To clearly understand this wiki, you should already know some elementary methods of breaking a rational function into i
appropriate partial fractions. The cover-up rule applies to computing coe cients of linear terms when the denominator
partial fraction decomposition consists of linear factors only or a combination of linear and irreducible factors.
Contents
Introduction
Applications
Problem Solving
See Also
Introduction
To compute the coe cients using the cover-up method, first set up a partial fraction decomposition with one term for ea
the factors in the denominator. For example, if the denominator has three distinct linear terms, we have the decomposit
f (x)
A
B
C
=
+
+
.
(x − a)(x − b)(x − c)
x−a x−b x−c
Then by the cover-up method, A can be computed by covering up the term (x − a) in the denominator of the le -hand
and substituting x = a in the remaining expression. This works because the computation is equivalent to multiplying th
expression throughout by the term (x − a) and then making the substitution x = a. This gives
A=
f (a)
.
(a − b)(a − c)
Similarly, by substituting x = b and x = c, we can compute B and C :
B=
f (b)
,
(b − a)(b − c)
C=
f (c)
.
(c − a)(c − b)
Note: Keep in mind that in order to apply partial fractions, the degree of the polynomial in the numerator must be stric
smaller than the degree of the polynomial in the denominator. If this is not the case, then it is necessary to first apply
polynomial division to obtain a quotient polynomial and a remainder, for which the degree of the numerator is strictly
smaller than that of the denominator. Partial fractions may then be applied to the remainder.
Here is a basic example on how to use the partial fraction rule for factorization.
EXAMPLE
Given the partial fraction
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3x
A
B
=
+
,
(x − 1)(x + 2)
x−1 x+2
what is the value of A + B?
Note that this partial fraction has two distinct linear factors. To obtain A, cover up the factor (x − 1) on the le -hand s
and substitute x = 1 into the remaining terms to obtain
A=
3(1)
3
= = 1.
1+2
3
3x
Similarly, to compute B , substitute x = −2 into x−1
to get
B=
3(−2)
−6
=
= 2.
−2 − 1
−3
Hence, A + B = 1 + 2 = 3. □
Try the following problem based on the understanding of applying partial fraction.
TRY IT YOURSELF
x2 − P
A
B
C
=
+
+
(x − 2)(x − 3)(x − 5)
x−2 x−3 x−5
Submit your answer
The equation above represents a partial fraction decomposition for constants A, B, C and
P.
What is the smallest value of the prime number P such that A, B and C are all integers?
TRY IT YOURSELF
Given that
Submit your answer
1−
1 1 1
π
+ − +⋯= ,
3 5 7
4
find the value of n that satisfy the equation below:
1
1
1
π
+
+
+⋯= .
1 × 3 5 × 7 9 × 11
n
Applications
This section is associated to the scenarios where the cover-up rule can be applied. Here are a couple of examples followe
some problems based on the applications of the rule:
EXAMPLE
Find the partial factor decomposition of
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x2 + x − 1
.
x(x2 − 1)
Observe that the denominator is x(x2 − 1) = x(x − 1)(x + 1), which is a product of distinct linear factors. Then the
partial factor decomposition can be wri en as
x2 + x − 1
A
B
C
=
+
+
.
x(x + 1)(x − 1)
x
x+1 x−1
To compute A, cover up the term x in the denominator of the le hand side and substitute x = 0 in the remaining ter
2
0 +0−1
obtain A = (0+1)(0−1)
= −1
= 1. Similarly, cover up (x + 1) and substitute x = −1 to obtain B = −1
and cover up
−1
2
(x − 1) and substitute x = 1 to obtain C = 12 . Therefore, the partial fraction decomposition is
x2 + x − 1
1
1
1
= −
+
.□
2
x(x − 1)
x 2(x + 1) 2(x − 1)
EXAMPLE
Given the partial fraction
6x
Ax + B
C
=
+
,
(x2 + 2)(x − 1)
x2 + 2
x−1
what is the value of A + B + C?
This partial fraction decomposition has one irreducible quadratic factor and one linear factor. We calculate the linear te
coe cient C by covering up the term x − 1 in the le -hand side and substitute x = 1 into the remaining terms. This
C = 126+2 = 2. To calculate A and B , we substitute this value of C into the equation to obtain
6x
Ax + B
2
+
2
x +2
x−1
6x
2
Ax + B
−
= 2
2
(x + 2)(x − 1) x − 1
x +2
2
6x − 2(x + 2)
Ax + B
= 2
2
(x + 2)(x − 1)
x +2
−2(x − 1)(x − 2)
Ax + B
= 2
2
(x + 2)(x − 1)
x +2
−2(x − 2)
Ax + B
= 2
.
2
(x + 2)
x +2
(x2 + 2)(x − 1)
=
Therefore, A = −2, B = 4, which gives A + B + C = −2 + 4 + 2 = 4. □
This problem checks the understanding of the usage of cover-up rule in factorial terms.
TRY IT YOURSELF
1
2
3
4
+ + + +⋯=?
2! 3! 4! 5!
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Submit your answer
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The long equations in the following two problems can be easily reduced to a shorter form by applying the partial fraction
TRY IT YOURSELF
Find the positive integer x such that
Submit your answer
1
1
1
1
= 2
+ 2
+ 2
.
10x
x + x x + 3x + 2
x + 5x + 6
TRY IT YOURSELF
1
1
1
1
+
+
=
(x − 1)(x − 2) (x − 2)(x − 3) (x − 3)(x − 4)
6
Submit your answer
What is the sum of all real values of x that satisfy the above equation?
Further problems also require wise usage of partial fraction rule to tackle them easily.
TRY IT YOURSELF
1
1
1
1
1
+
+
+
+
⋯
+
22 − 1 42 − 1 62 − 1 82 − 1
10002 − 1
Submit your answer
The value of the sum above can be expressed as ab , where a and b are coprime positive
integers. Find the value of a + b.
TRY IT YOURSELF
Evaluate
Submit your answer
∞
∑
n=0
n
n4 + n2 + 1
.
Note: Refrain from using Wolfram Alpha to solve this problem.
Problem Solving
This section contains several problems trying to build problem-solving skills involving the usage of the partial fraction ru
TRY IT YOURSELF
1
2
3
99
x = 101! × ( + + + ⋯ +
)
2! 3! 4!
100!
Submit your answer
Find 101! − x.
TRY IT YOURSELF
The sum
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Submit your answer
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1
1
1
1
+
+
+⋯+
2 1+1 2
3 2+2 3
4 3+3 4
100 99 + 99 100
a
can be expressed as b , where a and b are coprime positive integers.
What is the value of a + b?
This problem is adapted from a past year KMC Contest.
TRY IT YOURSELF
∞
1
1
1
1
1
∑ 2
=
+
+
+
+⋯
2
(n
−
1)n
3
⋅
4
8
⋅
9
15
⋅
16
24
⋅
25
n=2
The series above is equal to
and B coprime.
Submit your answer
A − Bπ C
where A, B, C and D are positive integers with A
D
Find the value of A + B + C + D .
TRY IT YOURSELF
∞
1
, what is the value of A1 ?
n(n
+
1)(n
+
2)(n
+
3)(n
+
4)(n
+
5)
n=1
If A = ∑
Submit your answer
TRY IT YOURSELF
∞
1
π2
∑ 2 =
,
k
6
k=1
∞
Si = ∑
Given the above, S1 + S2 can be represented as
k=1 (36k
i
Submit your answer
2 − 1)i
π2
b
− .
a
c
Find the value of a + b + c with c a prime number.
See Also
Factorization of Polynomials
Method of Di erences
Telescoping Series
Polynomial Interpolation
Cite as: Partial Fractions - Cover Up Rule. Brilliant.org. Retrieved 23:11, July 10, 2020, from h ps://brilliant.org/wiki/partial-fractions-cover-up-rule/
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