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Partial Fraction Decomposition notes with exercise and solutions
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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 x + b 1
a 2 x + b 2
Ak akx + bk Example (be sure you could have found the constants): 3 x (x − 1)(x − 2)
x − 2
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
(a 1 x + b 1 )^2
Br (a 1 x + b 1 )r Example: 3 x + 1 (x − 1)^2 (x − 2)
x − 1
(x − 1)^2
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)
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
x − 1
x + 2
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
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