PARTIAL FRACTION
If a given fraction of factorization denomination, it is possible to write such as sum of fraction with those factors appearing in their denomination
e.g.
Partial fraction eases computation of rather complicated algebraic notation and it is also useful in expanding rationals as power of constitute variable
DECOMPOSITION OF F(x)/G(x) INTO PARTIAL FRACTION
1. DISTINCT LINEAR FACTOR: – Such that the denominator can be factorized into linear factors
Example
Resolve into partial fraction
Solution
Multiplying both sides of thus equation by the least common denominator (x +2) (x – 5), leads to the basic equation
3x + 2 = A(x – 5) + B(x + 2)
we can substitute any convenient values of x which will help determine the constant A and B values of x that are especially convenient are ones that make the factor (x + 2) and (x –5) equal to zero for instance, let x = –2, Then
3(–2) + 2 = A( –2–5 ) + 3 ( –2+2)
–6+ 2 = –7 A
–4 = – 7 A
A =
To solve for B, let x = 5
3(5) + 2 = A(5 – 5) +B(5 + 2)
17 = 7B
Example
Resolve into partial fraction
Solution
To solve for B, let x = –3
–32 – 4 (–3) + 5 = 12B
26 = 12B
To solve for C, let x = 3
(3)2 – 4(3) + 5 = 24C
9 – 12 + 5 = 24C
C =
Hence
Example
Resolve into partial fraction
COVER – UP METHOD
Another method that is used for decomposition (resolution) into partial fraction is called the cover – up method. It is applicable only when the denominator has linear factors
Example
To solve for A, let x = –3
Similarly, we find B by writing
To find B , write the fraction
To solve for B, let x =7
Note:
One can also use the method of comparing coefficients of terms to resolve into partial fraction
Example: Decompose into partial fraction
Solution
1–2x = A(x–3) + B(x+5)
1–2 x = Ax – 3A + Bx + 5B
Equating coefficients of x terms
A +B = –2 ……….(i)
Equating constant terms
–3A + 5B =1
Solving equation (1) and (2) simultaneously
In case it is an improper fraction
If is an improper fraction. The first step is divide the numerator by the denominator, so that the degree of denominator will be greater than the degree of the numerator
Example
Decompose into partial fraction
Solution
x
–1
x2 – 1 x3 – x2 – 4
x3 – x
– x2 + x – 4
– x2 + 1
x – 5
Solution
Example
2. Distinct linear and quadratic factor (ax2 + bx + c) which has no linear factor
Example : Resolve into partial fraction
Solution
=
2x2 – 3x + 1 = A (x2 + 5x + 2) + (Bx + C) (x + 1) – – – (1)
To solve for A, let x = –1
2(–1)2 – 3(–1) + 1 = A ((–1)2 + 5(–1)+2)
2 +3 + 1 = –2A
A= −3
2x2 – 3x +1 = –3x2 – 15x – 6 + Bx2 + Bx + Cx + C {expanding eqn (1) and substituting A= −3}
Equating coefficients of x2 term
B – 3 = 2
B = 5
Equating coefficient of x terms
B + C – 15 = – 3
C = –3 + 15 – B
C = –12 – B
C = 12 – 5 = 7
Hence,
Example
Resolve into partial fraction
Solution
1 – x = (Ax + B)(x2–x–1) + (Cx + D)(2x2 + 3)
1 – x = Ax3 – Ax2 – Ax + Bx2 – Bx – B + 2Cx3 + 3cx +2Dx2 + 3D
Equating the coefficients of x2
A + 2C = 0 – – – – – – – – – (i)
Equating the coefficients of x2
–A+B+2D = 0 – – – –– – – – – (ii)
Equating the coefficient of x
Equating the constants
– B + 3D = 1 – – – – –– – – – – –(iv)
from (i) A = –2C – – – – – – – – – – – (v)
substitute A = –2C into (ii) and (iii)
2C + B + 2D = 0 – – – – – – – – –(vi)
5C – B = – 1 – – – – – – – – – – – –(vii)
Add (vi) and (vii) together
7C + 2D = – 1 – – – – – – – –– – –(viii)
from (vii) B = 5C + 1 – – – – – – –– – – – –(ix)
substitute B = 5C+1 into – – – – – – – – –(iv)
– (5C+1) + 3D = 1
5C–3D = – 2 – – – – –(x)
solving (viii) and (x) simultaneously
7C + 2D = – 1
5C– 3D = –2
C =
Substituting C = into (ix)
B =
Also A = −2C =
Hence
Example
Resolve into partial fraction
Solution
Using long–division
4
x3–1 4x3 – 19x2 + 19x + 6
4x3
– 4
– 19x2 + 19x + 10
3.) Corresponding to each repeated linear factor of the form (ax+b)n is a partial fraction of the form
Example
Solve into partial fraction
Solution
2x2 = A(x–2)2 + B(x–2) + C
2x2 = A(x2–4x+4) + B(x–2) + C
2x2 = Ax2 – 4Ax+4A + Bx – 2B + C
Comparing coefficients of like terms
x2 terms: A = 2
x terms: – 4A + B = 0 {substitute A = 2 into –4A + B = 0}
B = 8
Constant term: 4A – 2B + C = 0
C = 8
3 = Ax(x+2) + B(x+2) + Cx2
To solve for B, let x = 0
3 = 2B
B =
To solve for C, let x = –2
3 = 4C
C =
To solve for A let x = 1
3 = 3A + 3B + C – – – – (i)
Example
Express in partial fraction
Solution
2x2 + 39x + 12 = A(2x+1)(x–3)+B(x–3)+C(2x+1)2 – – – (i)
To solve for B, let x = {2x + 1= 0}
To solve for C let x = 3 {x – 3 = 0}
147 = 49C
C =
To solve for A, let x = 0 in eqn (i)
12 = –3A – 3B + C
Substitute B = – 2, C = into equation (ii)
12 = – 3A – 3(–2) +
3A = −6 + – 12
3A = – 9
A = –3
Hence
Also note or factor of the form (ax2+bx+c)n, the partial fraction decomposition include the following sum of n fractions
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