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

1.7 Partial fractions 43
Solution Thedenominatorhasalreadybeenfactorized.Thelinearfactor,x−4,producesapartial
fraction of the form A
x −4 .
The quadratic factor,x 2 +6x +13, will not factorize further into two linear factors.
Thus this factor generates a partialfraction of the form
Bx+C
x 2 +6x+13 . Hence
3x 2 +11x +14
(x −4)(x 2 +6x +13) = A
x −4 + Bx+C
x 2 +6x+13
Multiplying by (x −4) and (x 2 +6x +13) produces
3x 2 +11x +14 =A(x 2 +6x +13) + (Bx +C)(x −4) (1.11)
The constantsA,BandC can now be found.
Puttingx = 4 into Equation (1.11) gives
106 =A(53)
A = 2
Equating the coefficients ofx 2 gives
3=A+B
B = 1
Equating the constant termon both sides gives
Hence
14 =A(13) −4C
C = 3
3x 2 +11x +14
x 3 +2x 2 −11x−52 = 2
x −4 + x +3
x 2 +6x+13
1.7.4 Improperfractions
The techniques of calculating partial fractions in Sections 1.7.1 to 1.7.3 have all been
applied to proper fractions. We now look at the calculation of partial fractions of improper
fractions. The techniques described in Sections 1.7.1 to 1.7.3 are all applicable
to improper fractions. However, when calculating the partial fractions of an improper
fraction, an extra term needs to be included. The extra term is a polynomial of degree
n − d, where n is the degree of the numerator and d is the degree of the denominator.
A polynomial of degree 0 is a constant, a polynomial of degree 1 has the form
Ax +B, a polynomial of degree 2 has the form Ax 2 +Bx +C, and so on. For example,ifthenumeratorhasdegree3andthedenominatorhasdegree2,thepartialfractions
will include a polynomial of degree n −d = 3 − 2 = 1, that is a term of the form
Ax +B. If the numerator and denominator are of the same degree, the fraction is improper.
The partial fractions will include a polynomial of degree n −d = 0, that is a
constantterm.