- Proper and improper rational fractions

- Property of a improper rational fraction.

- Partial fractions

- Partial fraction decomposition

- Exercises

If the degree of N(x) is less then the degree of D(x) the fraction is said to be a proper rational fraction, otherwise we call it an improper rational fraction.

N(x) rest(x) ---- = Quotient(x) + -------- D(x) D(x)Example

3x^{2}+ 4x 7 ---------- = (3x + 7) + --------- x - 1 x - 1

A A ------- ; --------- ; (x - a) (x - a)Examples^{n}Ax + B Ax + B -------------- ; ----------------- with b^{2}- 4c < 0 x^{2}+ bx + c (x^{2}+ bx + c)^{n}Here A, B, b and c are real constants.

5 --------- (x -2)^{2}2x + 3 ------------ x^{2}+ x+ 4 2x + 3 ---------------- (x^{2}+ x+ 4)^{2}

Method:

- Factor the denominator as far as possible in real factors of first and second degree.

Divide numerator and denominator by an appropriate real number so that, the denominator has only factors of the form(x - a)

^{n}and/or (x^{2}+ bx + c)^{n} -
The proper rational fraction can be written as a sum of partial fractions. The nature and the number of partial fractions depends on the nature of the factors in the denominator.

Each factor (x - a)

^{n}in the denominator causes a sum of n partial fractions A B C L ------- + ------- + ------- + ... + ------- 2 3 n (x - a) (x - a) (x - a) (x - a) Each factor (x^{2}+ bx + c)^{n}in the denominator causes the sum of n partial fractions Ax + B Cx + D Lx + M --------------- + --------------- + ...+ --------------- 2 2 2 2 n (x + bx + c) (x + bx + c) (x + bx + c) - unknown numerators are calculated by the method of indefinite coefficients. (See examples).

2x + 5 -------------------- x^{3}- 3x^{2}- 4x + 12 We factor the denominator x^{3}- 3x^{2}- 4x + 12 = (x-3)(x+2)(x-2) Each factor causes exactly 1 elementary fraction of the sum 2x + 5 A B C ---------------- = ------- + ------- + ------- (x-3)(x+2)(x-2) (x - 3) (x + 2) (x - 2) A, B and C are presently unknown. Now we'll show how to calculate A, B and C. First we write the right side with one common denominator. Then the denominators are equal on both sides. So, the numerators must be equal. (2x + 5) = A(x + 2)(x - 2) + B(x - 3)(x - 2) + C(x - 3)(x + 2) (2x + 5) = (A + B + C)x.x + (-5B - C)x + (-4A +6B -6C) <=> / A + B + C = 0 | -5B - C = 2 \ -4A +6B -6C=5 <=> . . . <=> A = 2.2 B = 0.05 C = -2.25 Conclusion: 2x + 5 2.2 0.05 2.25 ---------------- = ------- + ------- - ------- (x-3)(x+2)(x-2) (x - 3) (x + 2) (x - 2)

7x --------------------- x^{3}- 6x^{2}+ 12x - 8 We factor the denominator and we find (x - 2)^{3}This factor causes a sum of three partial fractions 7x A B C -------- = ------- + ------- + ------- 3 2 3 (x - 2) (x - 2) (x - 2) (x - 2) A, B and C are presently unknown. Now we'll show how to calculate A, B and C. First we write the right side with one common denominator. Then the denominators are equal on both sides. So, the numerators must be equal. 7x = A (x - 2)^{2}+ B (x-2) + C <=> 7x = A x^{2}+ (B - 4A) x + 4A - 2B + C with the same method as in previous example we calculate A, B and C. A = 0 ; B = 7; C = 14 Conclusion: 7x 7 14 -------- = ------- + ------- 3 2 3 (x - 2) (x - 2) (x - 2)

2 2 5x + 4x + 1 5x + 4x + 1 ----------------- = ----------------- 3 2 2 2 3 (x + x )(x + 1) (x )(x + 1) The factor x^{2}in the denominator causes a sum of two partial fractions. The factor (x + 1)^{3}in the denominator causes a sum of three partial fractions. A B C D E --- + ---- + ------- + ------- + ------- 2 2 3 x x (x + 1) (x + 1) (x + 1) A, B, C, D and E are presently unknown. Now we'll show how to calculate A, B, C, D and E First we write the right side with one common denominator. Then the denominators are equal on both sides. So, the numerators must be equal. 5 x^{2}+ 4x + 1 = A x (x + 1)^{3}+ B (x + 1)^{3}+ C x^{2}(x + 1)^{2}+ D x^{2}(x + 1) + E x^{2}<=> 5 x^{2}+ 4x + 1 =(A + C) x^{4}+ (3 A + B + 2C + D)x^{3}+ (3A + 3B + C + D + E)x^{2}+ (A + 3B)x + B <=> / A + C = 0 | 3 A + B + 2C + D = 0 | 3 A + 3B + C + D + E = 5 | A + 3B = 4 \ B = 1 <=> ... <=> A = 1 ; B = 1; C = -1 ; D = -2 ; E = 2 Conclusion: 1 1 -1 -2 2 --- + ---- + ------- + ------- + ------- 2 2 3 x x (x + 1) (x + 1) (x + 1)

(2x + 1) ------------------------ 2 2 ( x + 1)( x + x + 1) The denominator can't be factored in real factors of lower degree. Each factor in the denominator gives us just one elementary fraction with a numerator of the first degree. (2x + 1) Ax + B Cx + D ---------------------- = --------------- + ----------------- 2 2 2 2 ( x + 1)( x + x + 1) ( x + 1) ( x + x + 1) A, B, C and D are presently unknown. Now we'll show how to calculate A, B, C, and D. First we write the right side with one common denominator. Then the denominators are equal on both sides. So, the numerators must be equal. (2x + 1) = (Ax + B)(x^{2}+ x + 1) + (Cx + D)(x^{2}+ 1) <=> (2x + 1) = (A + C) x^{3}+ (A + B + D) x^{2}+ (A + B + C) x + B + D and with the aid of a system with 4 unknowns, we find ... A = -1 ; B = 2 ; C = 1; D = -1 Conclusion (2x + 1) - x + 2 x - 1 ---------------------- = --------------- + ----------------- 2 2 2 2 ( x + 1)( x + x + 1) ( x + 1) ( x + x + 1)

x^{4}+ 5x^{3}+ 16x^{2}+ 26x + 22 ------------------------------ x^{3}+ 3 x^{2}+ 7x + 5 This improper fraction can be written as the sum of a polynomial and a proper rational fraction. 3 x^{2}+ 7 x + 12 (x+2) + ------------------------- x^{3}+ 3 x^{2}+ 7x + 5 We factor the denominator in (x+1)(x^{2}+ 2x + 5) 3 x^{2}+ 7 x + 12 A B x + C ---------------------- = ------------ + ----------------- (x+1)(x^{2}+ 2x + 5) x + 1 x^{2}+ 2x + 5 A, B and C are presently unknown. First we write the right side with one common denominator. Then the denominators are equal on both sides. So, the numerators must be equal. 3x2 + 7x + 12 = A(x^{2}+ 2x + 5) + (Bx + C)(x + 1) As above, you can find A, B and C by solving a system of three equations. A = 2 ; B = 1 ; C = 2 3 x^{2}+ 7 x + 12 2 x + 2 ---------------------- = ------------ + ----------------- (x+1)(x^{2}+ 2x + 5) x + 1 x^{2}+ 2x + 5

16 1 1 4 ------------------ = ----------- - ----------- - ------------ x^{3}- x^{2}- 5x - 3 (x - 3) (x + 1) (x + 1)^{2}- 3 x^{3}+ 8 x^{2}- 4 x + 5 x 3 1 ---------------------------------- = ------------ + ------- - ------ - x^{4}+ 3 x^{3}- 3 x^{2}+ 3 x - 2 x^{2}+ 1 x - 1 x - 2 2 x^{3}+ 7 x^{2}- 2 x + 6 2x 3 ------------------------ = ------------------ + --------------- x^{4}+ 4 x^{2}- 2 x + 2 x^{2}+ 2 x + 2

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