University:
Whitman College
Course:
MATH-124 | Introduction to Calculus
Academic year:

2023
Views:

500

Pages:

3
Author:

Reagan Huang

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) = (a1 x + b1 )(a2 x + b2 ) . . . (ak x + bk ) Then: P (x) A1 A2 Ak = + + ... + Q(x) a1 x + b 1 a2 x + b 2 ak x + b k Example (be sure you could have found the constants): 3x A B 6 3 = + = − (x − 1)(x − 2) x−2 x−1 x−2 x−1 Case II : Q has some repeated linear factors. Let a1 x + b1 be repeated r times. Then, instead of the single term A1 /(a1 x + b1 ), we have one term for each successive power in the denominator: B2 Br B1 + + ... + 2 a1 x + b1 (a1 x + b1 ) (a1 x + b1 )r Example: A B C 7 4 7 3x + 1 = + + =− − + 2 2 2 (x − 1) (x − 2) x − 1 (x − 1) x−2 x − 1 (x − 1) x−2 Case III : Q has some irreducible quadratic factors, not repeated. Let ax2 + bx + c be an irreducible quadratic factor for Q. Then the decomposition will have the term: Ax + B ax2 + bx + c Example: 3x A Bx + C 3 1 1 −3x + 3 = + 2 = + 2 (x − 1)(x + 1) x−1 x +1 2 x − 1 2 x2 + 1 Case IV : Q has some irreducible quadratic factors, some repeated. Suppose that ax2 + bx + c is a repeated quadratic factor (repeated r times). Then, instead of the single expression in Case III, we will have: A1 x + B1 A2 x + B2 Ar x + Br + + . . . + ax2 + bx + c (ax2 + bx + c)2 (ax2 + bx + c)r Example: x+3 1 x+1 2x + 1 = − 2 − 2 2 2 (x − 1)(x + 1) x − 1 x + 1 (x + 1)2 1 Worked Examples and Exercises 1. Worked Example: 6x2 − 6x − 6 A B C = + + (x − 1)(x + 2)(x − 3) x−1 x+2 x−3 Clear fractions: 6x2 − 6x − 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 2. Worked Example: A Bx + C x2 − 2 = + 2 x(x2 + 2) x x +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: x2 − 2 = A(x2 + 2) + (Bx + C)x = (A + B)x2 + Cx + 2A so that: 1 = A + B, 0 = C, 2A = −2 so: A = −1, B = 2 and C = 0: x2 − 2 −1 2x = + 2 2 x(x + 2) x x +2 Exercises For each of the following, first give the general form for the Partial Fraction expansion, then solve for the constants. x2 + 1 1. 2 x + 3x + 2 2. 2x + 3 (x + 1)2 3. 4x2 − 7x − 12 x(x + 2)(x − 3) 4. x2 + 3 x3 + 2x 5. x2 − 2x − 1 (x − 1)2 (x2 + 1) 6. 3x3 − x + 12 x2 − 1 2 Solutions 1. x2 + 1 5 2 =1− + 2 x + 3x + 2 x+2 x+1 2. 2x + 3 1 2 + = 2 (x + 1) x + 1 (x + 1)2 3. 2 9 1 1 1 4x2 − 7x − 12 = + + x(x + 2)(x − 3) x 5 x+2 5 x−3 4. x2 + 3 3 1 1 x = − 3 2 x + 2x 2 x 2 x +2 5. x2 − 2x − 1 1 1 x−1 = − − 2 2 2 2 (x − 1) (x + 1) x − 1 (x − 1) x +1 6. 3x + 5 7 − x−1 x+1 3

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