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Partial Fraction Decomposition (Summary), Lecture notes of Calculus

Partial Fraction Decomposition notes with exercise and solutions

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Partial Fraction Decomposition (Summary)
Partial Fraction Decomposition is used when we have a fraction, P(x)/Q(x), where P, Q are
polynomials, and the degree of Pis less than the degree of Q.NOTE: If the degree of the
numerator is larger than the denominator, then perform long division first.
Assume Qis fully factored. We have 4 cases that we will consider.
Case I :Qhas distinct linear factors,
Q(x) = (a1x+b1)(a2x+b2). . . (akx+bk)
Then: P(x)
Q(x)=A1
a1x+b1
+A2
a2x+b2
+. . . +Ak
akx+bk
Example (be sure you could have found the constants):
3x
(x1)(x2) =A
x2+B
x1=6
x2
3
x1
Case II :Qhas some repeated linear factors. Let a1x+b1be repeated rtimes. Then, instead
of the single term A1/(a1x+b1), we have one term for each successive power in the
denominator: B1
a1x+b1
+B2
(a1x+b1)2+. . . +Br
(a1x+b1)r
Example:
3x+ 1
(x1)2(x2) =A
x1+B
(x1)2+C
x2=
7
x1
4
(x1)2+7
x2
Case III :Qhas some irreducible quadratic factors, not repeated. Let ax2+bx +cbe an
irreducible quadratic factor for Q. Then the decomposition will have the term:
Ax +B
ax2+bx +c
Example: 3x
(x1)(x2+ 1) =A
x1+Bx +C
x2+ 1 =3
2
1
x1+1
2
3x+ 3
x2+ 1
Case IV :Qhas some irreducible quadratic factors, some repeated. Suppose that ax2+bx +cis
a repeated quadratic factor (repeated rtimes). Then, instead of the single expression
in Case III, we will have:
A1x+B1
ax2+bx +c+A2x+B2
(ax2+bx +c)2+. . . +Arx+Br
(ax2+bx +c)r
Example: x+ 3
(x1)(x2+ 1)2=1
x1
x+ 1
x2+ 1
2x+ 1
(x2+ 1)2
1
pf3

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Partial Fraction Decomposition (Summary)

Partial Fraction Decomposition is used when we have a fraction, P (x)/Q(x), where P, Q are polynomials, and the degree of P is less than the degree of Q. NOTE: If the degree of the numerator is larger than the denominator, then perform long division first.

Assume Q is fully factored. We have 4 cases that we will consider. Case I : Q has distinct linear factors, Q(x) = (a 1 x + b 1 )(a 2 x + b 2 )... (akx + bk) Then: P (x) Q(x)

A 1

a 1 x + b 1

A 2

a 2 x + b 2

Ak akx + bk Example (be sure you could have found the constants): 3 x (x − 1)(x − 2)

A

x − 2

B

x − 1

x − 2

x − 1

Case II : Q has some repeated linear factors. Let a 1 x + b 1 be repeated r times. Then, instead of the single term A 1 /(a 1 x + b 1 ), we have one term for each successive power in the denominator: B 1 a 1 x + b 1

B 2

(a 1 x + b 1 )^2

Br (a 1 x + b 1 )r Example: 3 x + 1 (x − 1)^2 (x − 2)

A

x − 1

B

(x − 1)^2

C

x − 2

x − 1

(x − 1)^2

x − 2

Case III : Q has some irreducible quadratic factors, not repeated. Let ax^2 + bx + c be an irreducible quadratic factor for Q. Then the decomposition will have the term: Ax + B ax^2 + bx + c

Example: 3 x (x − 1)(x^2 + 1)

A

x − 1

Bx + C x^2 + 1

x − 1

− 3 x + 3 x^2 + 1 Case IV : Q has some irreducible quadratic factors, some repeated. Suppose that ax^2 + bx + c is a repeated quadratic factor (repeated r times). Then, instead of the single expression in Case III, we will have: A 1 x + B 1 ax^2 + bx + c

A 2 x + B 2 (ax^2 + bx + c)^2

Arx + Br (ax^2 + bx + c)r

Example: x + 3 (x − 1)(x^2 + 1)^2

x − 1

x + 1 x^2 + 1

2 x + 1 (x^2 + 1)^2

Worked Examples and Exercises

  1. Worked Example: 6 x^2 − 6 x − 6 (x − 1)(x + 2)(x − 3)

A

x − 1

B

x + 2

C

x − 3 Clear fractions: 6 x^2 − 6 x − 6 = A(x + 2)(x − 3) + B(x − 1)(x − 3) + C(x − 1)(x + 2) And this statement must be true for all x. In particular, it must be true for x = 1, x = − 2 and x = 3 (we chose these to zero out the others). Substituting, we get A = 1 B = 2 C = 3

  1. Worked Example: x^2 − 2 x(x^2 + 2)

A

x

Bx + C x^2 + 2 Clear fractions. In this case, it might be best to solve for the coefficients in a slightly different manner- Equate the coefficients to the polynomials on the left and right: x^2 − 2 = A(x^2 + 2) + (Bx + C)x = (A + B)x^2 + Cx + 2A so that: 1 = A + B, 0 = C, 2 A = − 2 so: A = − 1 , B = 2 and C = 0: x^2 − 2 x(x^2 + 2)

x

2 x x^2 + 2

Exercises

For each of the following, first give the general form for the Partial Fraction expansion, then solve for the constants.

x^2 + 1 x^2 + 3x + 2

2 x + 3 (x + 1)^2

4 x^2 − 7 x − 12 x(x + 2)(x − 3)

x^2 + 3 x^3 + 2x

x^2 − 2 x − 1 (x − 1)^2 (x^2 + 1)

3 x^3 − x + 12 x^2 − 1