Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.
A General Note: Partial Fraction Decomposition of Q(x)P(x):Q(x) Has Repeated Linear Factors
The partial fraction decomposition of
Q(x)P(x), when
Q(x) has a repeated linear factor occurring
n times and the degree of
P(x) is less than the degree of
Q(x), is
Q(x)P(x)=(ax+b)A1+(ax+b)2A2+(ax+b)3A3+⋅⋅⋅+(ax+b)nAn
Write the denominator powers in increasing order.
Example 2: Decomposing with Repeated Linear Factors
Decompose the given rational expression with repeated linear factors.
x3−4x2+4x−x2+2x+4
Solution
The denominator factors are
x(x−2)2. To allow for the repeated factor of
(x−2), the decomposition will include three denominators:
x,(x−2), and
(x−2)2. Thus,
x3−4x2+4x−x2+2x+4=xA+(x−2)B+(x−2)2C
Next, we multiply both sides by the common denominator.
x(x−2)2[x(x−2)2−x2+2x+4]=[xA+(x−2)B+(x−2)2C]x(x−2)2 −x2+2x+4=A(x−2)2+Bx(x−2)+Cx
On the right side of the equation, we expand and collect like terms.
−x2+2x+4=A(x2−4x+4)+B(x2−2x)+Cx =Ax2−4Ax+4A+Bx2−2Bx+Cx =(A+B)x2+(−4A−2B+C)x+4A
Next, we compare the coefficients of both sides. This will give the system of equations in three variables:
−x2+2x+4=(A+B)x2+(−4A−2B+C)x+4A
A+B=−1−4A−2B+C=24A=4(1)(2)(3)
Solving for
A , we have
4A=4 A=1
Substitute
A=1 into equation (1).
A+B=−1(1)+B=−1 B=−2
Then, to solve for
C, substitute the values for
A and
B into equation (2).
−4A−2B+C=2−4(1)−2(−2)+C=2 −4+4+C=2 C=2
Thus,
x3−4x2+4x−x2+2x+4=x1−(x−2)2+(x−2)22