Partial Fractions

Question 1

• 2x - 5 / ( 4 + x ) ( 2 - x ) = A / ( 4 + x ) + B / ( 2 - x )

Answer 1

• ( First of all, identify whether the fraction is improper or proper, in this case it’s proper )
• ( Then look at the denominators and see what values are missing on the RHS compared to the LHS )
• • 2x - 5 / ( 4 + x ) ( 2 - x ) = A ( 2 - x ) + B ( 4 + x ) / ( 4 + x ) ( 2 - x )
• ( Denominators cancel out )
• • 2x - 5 = A ( 2 - x ) + B ( 4 + x )
• ( To work out values of A and B, you must substitute values of X that can cancel out that particular bracket )
• If x = 2
• • 2 ( 2 ) - 5 = A ( 2 - 2 ) + B ( 4 + 2 )
• • 9 = 6B
• B = - 9 / 2 = - 3 / 2
• If x = - 4
• • 2 ( - 4 ) - 5 = A ( 2 + 4 ) + B ( 4 - 4 )
• 3 = 6A
• A = 3 / 6 = 1 / 2
• 1 / 2 ( 4 + x ) - 3 / 2 ( 2 - x )
• ( Keep it in terms with what is shown to you in the question )

Question 2

Content on repeated roots:

Answer 2

3x^2 + x + 1 / x^2 ( x + 1 ) = A / x + B / x^2 + C / x + 1

- ( x^2 is split into x and x^2 terms, consecutive terms )

Question 3

Given that, for x < 0, 2x^2 + 2x - 18 / x ( x - 3 )^2 = P / x + Q / x - 3 + R / ( x - 3 )^2
Where P, Q and R are constants, find the values of P, Q and R.

Answer 3

• 2x^2 + 2x - 18 / x ( x - 3 )^2 = P / x + Q / x - 3 + R / ( x - 3 )^2
• 2x^2 + 2x - 18 / x ( x - 3 )^2 = P ( x - 3 )^2 + Q ( x ) ( x - 3 ) + R ( x ) / x ( x - 3 )^2
• ( Denominators cancel )
• If x = 3
• 2 ( 3 )^2 + 2 ( 3 ) - 18 = P ( 3 - 3 )^2 + Q ( 3 ) ( 3 - 3 ) + R ( 3 )
• 6 = 3R
• R = 2
• If x = 0
• 2 ( 0 )^2 + 2 ( 0 ) - 18 = P ( 0 - 3 )^2 + Q ( 0 ) ( 0 - 3 ) + R ( 0 )
• • 18 = 9P
• p = - 2
• If x = 1
• ( To find the last values, substitute values of the constants in order to find the last constant )
• 2 ( 1 )^2 + 2 ( 1 ) - 18 = - 2 ( 1 - 3 )^2 + Q ( 1 ) ( 1 - 3 ) + 2 ( 1 )
• • 14 = - 8 - 2Q + 2
• • 8 = - 2Q
• Q = 4

Question 4

Show that 39x^2 + 2x + 59 / ( x + 5 ) ( 3x - 1 )^2 can be written in the form A / x + 5 + B / 3x - 1 + C / ( 3x - 1 )^2 where A, B and C are constants to be found.

Answer 4

• 39x^2 + 2x + 59 / ( x + 5 ) ( 3x - 1 )^2 = A ( 3x - 1 )^2 + B ( 3x - 1 ) ( x + 5 ) + C ( x + 5 ) / ( x + 5 ) ( 3x - 1 )^2
• ( Denominators cancel out )
• 39x^2 + 2x + 59 = A ( 3x - 1 )^2 + B ( 3x - 1 ) ( x + 5 ) + C ( x + 5 )
• If x = - 5
• 39 ( - 5 )^2 + 2 ( - 5 ) + 59 = A ( 3 ( - 5 ) - 1 )^2 + B ( 3 ( - 5 ) - 1 ) ( - 5 + 5 ) + C ( - 5 + 5 )
• 1024 = 256A
• A = 4
• If x = 1 / 3
• 39 ( 1 / 3 )^2 + 2 ( 1 / 3 ) + 59 = A ( 3 ( 1 / 3 ) - 1 )^2 + B ( 3 ( 1 / 3 ) - 1 ) ( 1 / 3 + 5 ) + C ( 1 / 3 + 5 )
• 64 = 16 / 3C
• C = 12
• If x = 1
• ( Substitute constants )
• 39 ( 1 )^2 + 2 ( 1 ) + 59 = 4 ( 3 ( 1 ) - 1 )^2 + B ( 3 ( 1 ) - 1 ) ( 1 + 5 ) + 12 ( 1 + 5 )
• 100 = 16 + 12B + 72
• 100 = 12B + 88
• 12 = 12B
• B = 1

Question 5

6x^3 - 7x^2 + 3 / 3x^2 - x - 10 = Ax + B + C / 3x - 5 + D / x + 2

Answer 5

• ( This is an improper fraction, therefore requires long division first )
• ( Improper = long division )
• 3x^2 + x - 10 | 6x^3 - 7x^2 + 0x + 3 | ( Division )
• ( Division only gets you so far because the x values of the divisor cannot be bigger than the x values in the expression which is being divided )
• 2x - 3 = the Quotient ( The answer to the division )
• 23x - 27 = the remainder
• Ax + B = 2x - 3
• So A = 2 and B = - 3
• ( Use the remainder as the numerator and keep the denominator constant to find values of C and D )
• 23x - 27 / 3x^2 + x - 10 = C / 3x - 5 + D / x + 2
• ( factorise the denominator )
• 23x - 27 / ( 3x - 5 ) ( x + 2 ) = C ( x + 2 ) + D ( 3x - 5 ) / ( 3x - 5 ) ( x + 2 )
• ( Denominators cancel out )
• 23x - 27 = C ( x + 2 ) + D ( 3x - 5 )
• If x = - 2
• 23 ( - 2 ) - 27 = C ( - 2 + 2 ) + D ( 3 ( - 2 ) - 5 )
• • 73 = - 11D
• D = 73 / 11 = 6.64
• If x = 1 ( Substituting values of D )
• 23 ( 1 ) - 27 = C ( 1 + 2 ) + 73 / 11 ( 3 ( 1 ) - 5 )
• • 3 = 3C - 146 / 11
• 102 / 11 = 3C
• C = 34 / 11