082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

42 Chapter 1 Review of algebraic techniques
The factor, (x −1), produces a partialfraction ofthe form C
x −1 . So
14x 2 +13x
(4x 2 +4x +1)(x −1) = 14x 2 +13x
(2x +1) 2 (x −1) = A
2x+1 + B
(2x +1) + C
2 x −1
Multiplying both sides by (2x +1) 2 (x −1) gives
14x 2 +13x =A(2x +1)(x −1) +B(x −1) +C(2x +1) 2 (1.10)
The unknown constantsA,B andC can now befound.
Evaluating Equation (1.10) withx = 1 gives
27 =C(3) 2
from which
C = 3
Evaluating Equation (1.10) withx = −0.5 gives
14(−0.5) 2 +13(−0.5) =B(−0.5 −1)
from which
B = 2
Finally, comparing the coefficients ofx 2 on both sides of Equation (1.10) wehave
14=2A+4C
Since wealready haveC = 3 then
A = 1
Hence we see that
14x 2 +13x
(4x 2 +4x +1)(x −1) = 1
2x+1 + 2
(2x +1) 2 + 3
x −1
1.7.3 Quadraticfactors
We now look at proper fractions whose denominator contains a quadratic factor, that is
a factorofthe formax 2 +bx +c.
Aquadraticfactor,ax 2 +bx +c, producesapartialfraction ofthe form
Ax+B
ax 2 +bx+c
Example1.41 Notingthatx 3 +2x 2 −11x −52 = (x −4)(x 2 +6x +13), express
3x 2 +11x +14
x 3 +2x 2 −11x−52
as partialfractions.