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

40 Chapter 1 Review of algebraic techniques
Thelinearfactor,x +1,producesapartialfractionoftheform A
x +1 .Thelinearfactor,
x + 2, produces a partial fraction of the form B . A and B are unknown constants
x +2
whose values have tobefound. So we have
6x+8
x 2 +3x+2 = 6x+8
(x +1)(x +2) = A
x +1 + B
x +2
Multiplying both sides of Equation (1.8)by (x +1) and (x +2) weobtain
(1.8)
6x+8=A(x+2)+B(x+1) (1.9)
We now evaluateAandB. There are two techniques which enable us to do this: evaluation
using a specific value of xand equating coefficients. Each isillustratedinturn.
Evaluationusingaspecificvalueofx
We examine Equation (1.9). We will substitute a specific value ofxinto this equation.
Although any value can be substituted for x we will choose a value which simplifies
the equation as much as possible. We note that substitutingx = −2 will simplify the
r.h.s.oftheequationsincethetermA(x +2)willthenbezero.Similarly,substitutingin
x = −1 will simplify the r.h.s. because the termB(x +1) will then be zero. Sox = −1
andx = −2 are two convenient values to substitute into Equation (1.9). We substitute
each inturn.
Evaluating Equation (1.9) withx = −1 gives
−6+8=A(−1+2)
2 =A
Evaluating Equation (1.9) withx = −2 gives
−4 =B(−1)
B = 4
SubstitutingA = 2,B = 4into Equation (1.8) yields
6x+8
x 2 +3x+2 = 2
x +1 + 4
x +2
2
Thus the required partialfractions are
x +1 and 4
x +2 .
The constantsAandBcould have been found by equating coefficients.
Equatingcoefficients
Equation (1.9) may be writtenas
6x+8=(A+B)x+2A+B
Equating the coefficients ofxon both sides gives
6=A+B
Equating the constant termson both sides gives
8=2A+B