Partial Fraction, Advance Algebra, first-year college Mathematics, Bachelor of Secondary Mathematics Education

2. REQUIREMENTS OF BSED
MAJOR IN MATHEMATICS
PROJECT
FIRST SEMESTER 2018-2019
TITLE:PARTIAL FRACTION
STUDENT:CHRISTOPHER CAALIM
INSTRUCTOR:MATTHEW E. CAÑEDA
PRODUCER:Angelito alalan
DURATION:1 day

6. The partial fraction decomposition will contain an expression of the form
Partial fraction decomposition:
Each linear factor of the denominator occurs
only once
𝟑𝒙 − 𝟏
𝒙(𝟑𝒙 + 𝟒)(𝒙 − 𝟐)
𝟑𝒙 − 𝟏
𝒙(𝟑𝒙 + 𝟒)(𝒙 − 𝟐)
=
𝑨
𝒙
+
𝑩
(𝟑𝒙 + 𝟒)
+
𝑪
(𝒙 − 𝟐)
𝐴
(𝑎𝑥 + 𝑏) for each no repeated linear factor of the denominator.

7. The partial fraction decomposition will contain an expression
of the form.
(x-2)²=(x-2)(x-2)
a repeated linear factor
A1
(𝒂𝒙 + 𝒃)
+
A2
(𝒂𝒙 + 𝒃)²
+ ⋯ … +
A3
𝒂𝒙 + 𝒃 m
For each repeated linear factor of multiplicity m for example.
𝟒𝒙 + 𝟓
(𝒙 − 𝟐)²(𝟐𝒙 + 𝟏)
Partial fraction decomposition:
𝟒𝒙+𝟓
(𝒙−𝟐)²
=
A1
(𝒙−𝟐)
+
A2
(𝒙−𝟐)²
+
B
(𝟐𝒙+𝟏)

8. Case 3: Non repeated Quadratic Factors
The partial fraction decomposition will contain an expression
of the form
x²+x+1 is irreducible over the real number
𝑨𝒙 + 𝑩
𝒂𝒙 𝟐 + 𝒃𝒙 + 𝒄
For each quadratic factors that is irreducible over the real number for example:
𝒙 − 𝟒
(𝒙 𝟐 + 𝒙 + 𝟏)(𝒙 − 𝟒)
Decomposition:
𝒙−𝟒
(𝒙 𝟐+𝒙+𝟏)(𝒙−𝟒)
=
𝑨𝒙+𝑩
𝒙 𝟐+𝒙+𝟏
+
𝑪
𝒙−𝟒

9. The partial fraction decomposition will contain an expression of the
form
A1x+B1
𝒂𝒙 𝟐 + 𝒃𝒙 + 𝒄
+
A2x+B2
(𝒂𝒙 𝟐 + 𝒃𝒙 + 𝒄)²
+ ⋯ +
Amx+Bm
(𝒂𝒙 𝟐 + 𝒃𝒙 + 𝒄)m
For each quadratic factors that is irreducible over the real number for example:
𝟐𝒙
(𝒙 − 𝟐)(𝒙 𝟐 + 𝟒)²
=
A1x+B1
𝒙 𝟐 + 𝟒
+
A2x+B2
(𝒙 𝟐 + 𝟒)²
+
𝑪
𝒙 − 𝟐

10. Solution
First factor the denominator
Combine like terms
x²-2x-15=(x+3)(x-5)
The factors are non repeated linear factors. Therefor, the partial fraction decomposition
will have the form
𝒙 + 𝟏𝟏
(𝒙 + 𝟑)(𝒙 − 𝟓)
=
𝑨
𝒙 + 𝟑
+
𝑩
𝒙 − 𝟓
To solve for A and B, multiply each side of the equation by the least common
denominator multiple of the denominator, (x+3)(x-5)
x+11=A(x-5)+B(x+3)
x+11=(A+B)x+(-5A+3B)
Using the equality of polynomials theorem, equate coefficient of like
powers. The result will be the system of equations
1 = A+B
11=-5A+3B

11. A+B=1 B=1-A
-5A+3B=11
-5A+3(1-A)=11
-5A-3A+3=11
-8A=8
A=-1
B=1-(-1)
B=2
Solving the system of equations for A and B, we have A=-1 and B=2
substituting -1 for A and 2 for B into the form of the partial fraction
decomposition (case 1), we obtain
𝒙 + 𝟏𝟏
(𝒙 + 𝟑)(𝒙 − 𝟓)
=
−𝟏
𝒙 + 𝟑
+
𝟐
𝒙 − 𝟓

12. THANK YOU for watching!