Denominator: Linear and different factors Partial Fractions Type 1 Denominator: Linear and different factors Worked example

Find partial fractions for 5x – 11 2x2 + x - 6 Find partial fractions for 5x – 11 2x2 +x - 6 Check top line is lower degree 5x – 11 ( 2x – 3 )( x + 2 ) Factorise bottom line = Split into commonsense fractions. A and B are constants. Note: top line must be lower degree A ( 2x – 3 ) B ( x + 2 ) = + A( x + 2 ) + B( 2x – 3 ) ( 2x – 3 )( x + 2 ) Combine the fractions =

Now find the unknowns A and B. 5x – 11 2x2 +x - 6 A( x + 2 ) + B( 2x – 3 ) ( 2x – 3 )( x + 2 ) => = => 5x – 11 = Identity A( x + 2 ) + B( 2x – 3 ) Now find the unknowns A and B. There are two ways to do this. 1: Equate coefficients and solve a pair of simultaneous equations. 2: Substitute “clever” values of x into the identity i.e. values of x which make terms disappear We will use the second method.

Partial fractions are: Identity: 5x – 11 = A( x + 2 ) + B( 2x – 3 ) Let x = 3 2 Let x = -2 ( x + 2 = 0 ) ( 2x - 3 = 0 ) – 11 = A 7 2 15 -10 – 11 = -7B -21 = -7B B = 3 ……… x (2) 15 – 22 = 7A -7 = 7A A =-1 Partial fractions are: 5x – 11 2x2 +x - 6 -1 ( 2x – 3 ) 3 ( x + 2 ) = +

Denominator: Linear factor and an irreducible quadratic factor Partial Fractions Type 2 Denominator: Linear factor and an irreducible quadratic factor Worked example

Find partial fractions for 3x – 2 ( x + 1 )( x2 + 4 ) Find partial fractions for 3x – 2 ( x + 1 )( x2 + 4 ) Check top line is lower degree 3x – 2 ( x + 1 )( x2 + 4 ) Factorise denominator fully = Now ready to split into commonsense fractions. A, B and C are constants. Note: top line must be lower degree. A ( x + 1 ) Bx + C ( x2 + 4 ) = + A( x2 + 4 ) +( Bx + C )( x + 1 ) ( x + 1 )( x2 + 4 ) Combine the fractions =

Partial fractions are: Identity: 3x – 2 = A( x2 + 4 ) +( Bx + C )( x + 1 ) We have run out of clever values. Now choose simple values, not already used, and use the values of the constants already found Let x = -1 ( x + 1 = 0 ) -3 – 2 = A((-1)2 + 4) -5 = 5A A = -1 (but A = -1 and C = 2) Let x = 1 Let x = 0 –2 = 4A + C (but A = -1) 1 = 5A + ( B + C )( 2 ) -2 = -4 + C C = 2 1 = -5 + 2B + 4 2B = 2 B = 1 Partial fractions are: 3x – 2 ( x + 1 )( x2 + 4 ) -1 ( x + 1 ) x + 2 ( x2 + 4 ) = +

Denominator: Repeated linear factors Partial Fractions Type 3 Denominator: Repeated linear factors Worked example

Find partial fractions for 7x2 -11x - 5 ( x + 2 )( x - 1 )2 Find partial fractions for The factor ( x – 1 ) is repeated. There are 3 factors on the bottom line so we must split it into 3 bits. The correct way to split is as follows 7x2 -11x - 5 ( x + 2 )( x - 1 )2 A ( x + 2 ) B ( x - 1 ) C ( x - 1 )2 = + + Identity: 7x2 -11x - 5 = A( x - 1 )2 + B( x - 1 )( x + 2 ) + C ( x + 2 ) clever values Let x = 1 ( x - 1 = 0 ) Let x = -2 ( x + 2 = 0 ) 7 – 11 - 5 = C( 3 ) 28 + 22 - 5 = A( -3 )2 -9 = 3C 45 = 9A C = -3 A = 5

Partial fractions are: Identity: 7x2 -11x - 5 = A( x - 1 )2 + B( x - 1 )( x + 2 ) + C ( x + 2 ) We have run out of clever values. So choose simple values, not already used, and use the values of the constants already found. Let x = 0 –5 = A(-1)2 +B(-1)(2) + C(2) -5 = A – 2B + 2C ( but A = 5 and C = -3 ) -5 = 5 – 2B - 6 2B = 4 B = 2 Partial fractions are: 7x2 -11x - 5 ( x + 2 )( x - 1 )2 5 ( x + 2 ) 2 ( x - 1 ) 3 ( x - 1 )2 = + -

More splitting examples 3x2 + 2x + 9 ( x - 3 )( x - 2 )3 4 factors, so split into 4 bits A (x - 3 ) B ( x - 2 ) D ( x - 2 )3 = C ( x - 2 )2 + + + 5x + 7 x2( x + 1 ) x is repeated. 3 factors, so split into 3 bits A x B x2 C ( x + 1 ) = + + In general f(x) ( x + a )n n factors, so split into n bits A1 (x + a ) A2 (x + a )2 A3 (x + a )3 An (x + a )n = + + + . . . . . . +

Numerator same degree or higher than the denominator Partial Fractions Numerator same degree or higher than the denominator Worked example

Find partial fractions for x( x + 3 ) x2 + x - 12 Find partial fractions for Numerator is degree 2, denominator is degree 2. Same degree => DIVIDE OUT first 1 QUOTIENT x2 + x - 12 x2 + 3x + 0 x2 + x - 12 REMAINDER 2x + 12 NOW FIND PARTIAL FRACTIONS FOR THIS BIT x( x + 3 ) x2 + x - 12 2 x + 12 x2 + x - 12 => = + 1 THE FINAL ANSWER IS x( x + 3 ) x2 + x - 12 18 7( x – 3 ) 4 7( x + 4 ) = + - 1

Let’s do some more examples Express 3x2 + 2x + 1 ( x + 1 )(x2 + 2x + 2) in partial fractions. We must first use the discriminant to verify that the quadratic factor (x2 + 2x + 2) is irreducible. For x2 + 2x + 2 : a = , b = , c = 1 2 2 b2 – 4ac = 22 – 4(1)(2) = -4 b2 – 4ac < 0 => x2 + 2x + 2 is irreducible 3x2 + 2x + 1 ( x + 1 )(x2 + 2x + 2) =

as the sum of a polynomial and partial fractions. Another example: Express x3 - 3x (x2 - x - 2) as the sum of a polynomial and partial fractions.