At the end of the lesson, the students should be able to: (a) Construct partial fractions decomposition when the denominators in the form of (i) a linear factors, (ax + b) (ii) a repeated linear factors, (ax + b)n (iii) a quadratic factor, ax2 + bx + c that cannot be factorised OBJECTIVES :
A single proper algebraic fraction can be expressed as a sum or difference of two or more proper fractions. For example 3 1 1 2 1 1 x x x x x 2 x 1 1 x and are known as partial 3 1 1 x x x fractions of Partial Fractions
the degree of the numerator is greater than, or equal to, the degree of the denominator. Example: the degree of the numerator is less than the degree of the denominator. Example: There are two types of FRACTIONS frtions: 2 2 3 3 x 2x 7 or x 4 x 1 PROPER FRACTION IMPROPER FRACTION 3 3 2x 1 or x 5 4 2 2x 7 x 1
CASE 1 Proper fractions CASE 2 Improper fractions Long Division Partial Fractions The denominators are in the form of (i) a linear factors, (ax + b) (ii) a repeated linear factors, (ax + b)n (iii) a quadratic factor, ax2 + bx + c that cannot be factorised.
1 2 A B x x Example 2 3 ( 1)( 2) x x x The denominators are in the form of (i) a linear factors, (ax + b) The denominators are in the form of (ii) a repeated linear factors, (ax + b)n 2 2 1 2 1 x x x 2 2 1 A B C x x x repeated linear factors Example linear factors
The denominators are in the form of (iii) a quadratic factor, ax2 + bx + c that cannot be factorised Example 2 5 7 (2 1)( 1) x x x x 2 2 1 1 A Bx C x x x quadratic factors that cannot be factorised
1 2 A B x x 2 3 ( 2) ( 1) x A x B x Example 1 Express 2 3 ( 1)( 2) x x x as partial fractions. Solution 2 3 ( 1)( 2) x x x The denominators are in the form of (i) a linear factors, (ax + b)
2 3 (3) (0) A B When x = 1, 5 3A 3 5 A When x = - 2, 2( 2) 3 (0) ( 3) A B 3B 1 3 1 B Hence, 2 3 5 1 ( 2)( 1) 3( 1) 3( 2) x x x x x 2 3 ( 2) ( 1) x A x B x
Example 2 7 4 1 2 1 x x x x Express as partial fractions. Solution: 7 4 1 2 1 (1 ) (2 1) x A B C x x x x x x 7 4 (1 )(2 1) (2 1) (1 ) x A x x Bx x Cx x
When x = 0, When x = 1, When x = , 4 (1)(1) A 3 3(1) B A 4 B 1 C 10 7 4 (1 )(2 1) (2 1) (1 ) x A x x Bx x Cx x 1 2 7 4 4 1 10 1 2 1 (1 ) (2 1) x x x x x x x
Example 3 2 3 1 ( 1) x x x Express as partial fractions. Solution: 2 3 1 ( 1) x x x 2 3 1 ( 1) ( 1) x Ax x B x Cx 2 ( 1) A B C x x x
When x = 0, B = 1 When x = 1, C = 2 A 2 2 2 3 1 2 1 2 ( 1) ( 1) x x x x x x 3 = A + B Comparing the coefficient of x gives, 2 3 1 ( 1) ( 1) x Ax x B x Cx
Example 4 2 9 ( 1)( 2) x x Express in partial fractions Solution: ( )( ) 2 9 x x 1 2 ( ) ( )( ) ( ) 2 9 2 1 2 1 A x B x x C x( ) 2 1 2 2 A B C x x x
When x = 2, C = 3 When x = 1, A = 1 ( )( ) ( ) ( ) ( ) 2 2 9 1 1 3 x x x x x 1 2 1 2 2 Comparing the coefficients of x2 gives, 0 = A + B B = A = -1 ( ) ( )( ) ( ) 2 9 2 1 2 1 A x B x x C x
Example 5 2 4 1 3 x x x Express as partial fractions. 2 4 ( 1)(3 ) x x x Solution: 2 ( 1) (3 ) A Bx C x x 2 4 3 ( )( 1) x A x Bx C x
When x=-1, B 1 A 1 2 2 4 1 3 ( 1)(3 ) 3 1 x x x x x x 2 x A B : 0 x B C : 4 2 4 3 ( )( 1) x A x Bx C x Comparing the coefficients, C 3
Exercise 2 3 4 ( 2) 1 x x x x Express as a partial fractions Solution: 2 2 1 A Bx C x x x 2 3 4 1 ( )( 2) x A x x Bx C x 2 3 4 ( 2) 1 x x x x
2 3 4 1 ( )( 2) x A x x Bx C x When x = - 2, 2 7 A 2 7 B 2 2 3 4 2 2 15 ( 2)( 1) 7( 1) 7( 2) x x x x x x x x 2 x A B : 0, Comparing the coefficient, 15 7 When x = 0, A C 2 4 C
CASE 1 Proper fractions Partial Fractions The denominators are in the form of (i) a linear factors, (ax + b) (ii) a repeated linear factors, (ax + b)n (iii) a quadratic factor, ax2 + bx + c that cannot be factorised CONCLUSION