Partial Fractions

4038 ADDITIONAL MATHEMATICS TOPIC 1: ALGEBRA

SUB-TOPIC 1.5 PARTIAL FRACTIONS CONTENT OUTLINE 1. Solving cases whereby the denominator is no more complicated than: a. (ax + b)(cx + d) b. (ax + b)(cx + d)2 c. (ax + b)(x2 + c2)

4038 ADDITIONAL MATHEMATICS TOPIC 1: ALGEBRA SUB-TOPIC 1.5: PARTIAL FRACTIONS A Introduction The highest exponent is the degree of the polynomial. For example, 4x2 – 2x + 1 is a polynomial in x of degree 2, whereas 3x4 + 22x3 is a polynomial in x of degree 4. P(x)

If P(x) and Q(x) are two polynomials in x, then the fraction Q(x) is called an algebraic fraction. It is a proper algebraic fraction if the degree of P(x) is strictly lesser than that of Q(x). For 1

example, 13x + 3 ,

15x2

, 3x3 - 25

15(x+2)3 3x4+ 15

15x2

are all proper fractions, whereas 13x + 3 ,

15(x+2)3 3x3 - 25

,

x6 3x4+ 15

are

improper fractions. If an algebraic fraction is improper, we can express it into the sum of a polynomial and a proper fraction by long division or in some cases by suitable algebraic manipulation of the numerator so that more cancellation can be done. Assume the following fraction: fraction.

15x3+ 2x + 1 x+2

. Split it up into a sum of a polynomial and a mixed

Long division x+2

) -

Algebraic Manipulation 15x2 15x3 (15x3 -

- 30x + 0x2 + 30x2) - 30x2 (-30x2 -

+ 62 +2x + 2x - 60x) 62x (62x

= +1 +1 +1 + 124) - 123

= =

15x3 + 30x2 - 30x2 + 2x+1 x+2 15x2(x + 2) - 30x2 + 2x + 1

15x2

+

= 15x2 +

x+2 -30x2 - 60x + 60x + 2x + 1 x+2 -30x x + 2 + 62x + 1

= 15x2 – 30x +

x+2 62x + 1 x+2 62x + 124 – 124 + 1

=

15x2

– 30x +

=

15x2

– 30x +

=

15x2

– 30x + 62 -

x+2 62 x + 2 - 123 x+2 123 x+2

Sometimes, a fraction may have complicated denominators, such as (x + 2)(2x2 + 9). In such cases, it would be preferable to split up the fraction into two or more fractions with the denominators (x + 2) and (2x2 + 9). In the next few pages, we shall explore how to go about doing so.

4038 ADDITIONAL MATHEMATICS TOPIC 1: ALGEBRA SUB-TOPIC 1.5: PARTIAL FRACTIONS B

Linear Factors Fraction yx + z ax + b (cx + d)

Express

8x - 46 x - 5 (x + 1)

Partial Fraction A B + ax+ b cx + d

in partial fractions 8x - 46 x - 5 (x + 1) 8x – 46

[3]

= = =

A B + (x – 5) (x + 1) A(x + 1) + B(x – 5) Ax + A + Bx – 5B (A + B)x + (A – 5B)

Comparing coeffs of x term, 8 A

= =

A+B 8 – B --- (1)

Comparing coeffs of constant term, -46 A

= =

A – 5B 5B – 46 --- (2)

(1) 8–B 8 + 46 54 B

= = = = =

(2) 5B – 46 B + 5B 6B 9

Sub B = 9 into (1): A

= =

8–9 -1

Hence,

8x - 46 x - 5 (x + 1)

=

= =

-1 9 + (x – 5) (x + 1) 1 9 + (Ans) 5–x x+1 [N2008/I/3i]

4038 ADDITIONAL MATHEMATICS TOPIC 1: ALGEBRA SUB-TOPIC 1.5: PARTIAL FRACTIONS C

Repeated Factor Fraction yx + z

Partial Fraction A B C + + ax+ b cx+d (cx+d)2

ax + b (cx + d)2 Express in partial fractions,

9x2 2

x - 1 (x+2) 2

A B C 9x + + = (x + 2) (x – 1) (x - 1)2 x - 1 2 (x+2) 9x2 = A(x – 1)2 + B(x + 2)(x – 1) + C(x + 2) = A(x2 – 2x + 1) + B(x2 + x – 2) + C(x + 2) = Ax2 – 2Ax + A + Bx2 + Bx – 2B + Cx + 2C = (A + B)x2 + (-2A + B + C)x + (A – 2B + 2C) Comparing coeffs of x2 term, 9 = A + B A = 9 – B --- (1) Comparing coeffs of x term, 0 = -2A + B + C 2A = B + C --- (2) Comparing coeffs of constant term, 0 = A – 2B + 2C A = 2B – 2C --- (3) Sub 2(1) into (2): 2(9 – B) = B + C 18 – 2B = B + C 18 – 3B = C --- (4) Sub (1) into (3): 9 - B = 2B – 2C 9 – 3B = 2C --- (5) Sub -2(4) into (5): 9 – 3B 9 – 3B 9 + 36 45 B

= = = = =

-2 (18 – 3B) -36 + 6B 6B + 3B 9B 5

Sub B = 5 into (1): A = 9 – 5 = 4 Sub B = 5 into (4): C = 18 – 3 (5) = 18 – 15 = 3 Hence,

4 5 3 9x2 + + = (x + 2) (x – 1) (x – 1)2 x - 1 2 (x+2) [N2004/I/13 part] (AM)

4038 ADDITIONAL MATHEMATICS TOPIC 1: ALGEBRA SUB-TOPIC 1.5: PARTIAL FRACTIONS D

Quadratic Factor which does not Factorise Fraction yx + z ax + b (x2 + c2 )

Express in partial fractions,

17x2 + 23x+12 3x+4 (x2 + 4) 2

Partial Fraction A Bx + C + 2 2 ax + b x +c

.

17x + 23x+12 A Bx + C = + 2 2 3x + 4 (x + 4) 3x+4 x +4 17x2 + 23x + 12 = A(x2 + 4) + (Bx + C)(3x + 4) = Ax2 + 4A + 3Bx2 + 4Bx + 3Cx + 4C = (A + 3B)x2 + (4B + 3C)x + (4A + 4C) Compare coeffs of x2 term, 17 = A + 3B A = 17 – 3B --- (1) Compare coeffs of x term, 23 = 4B + 3C 4B = 23 – 3C --- (2) Compare coeffs of constant term, 12 = 4A + 4C 3 = A+C A = 3 – C --- (3) (1) 17 – 3B 17 – 3 14 C

= = = = =

Sub (4) into (2): 4B = = = 4B + 9B = 13B = B =

(3) 3–C 3B – C 3B – C 3B – 14 --- (4) 23 – 3(3B – 14) 23 – 9B + 42 65 – 9B 65 65 5

Sub B = 5 into (4): C = 3(5) – 14 = 15 – 14 = 1 Sub C = 1 into (3): A = 3 – 1 = 2 Hence,

17x2 + 23x+12 2 5x + 1 = + 2 2 3x + 4 (x + 4) 3x+4 x +4 [N99/I/16(a)] (AM)