Page 139 - Advanced engineering mathematics

3.7 Polynomial Coefficients 119

with the coefficients to be determined. One way to do this is to add the terms in S(x),set the
numerator of the resulting quotient equal to P(x), which is known, and solve for the coefficients
of the terms in S(x) by equating coefficients of like powers of x.

EXAMPLE 3.23
We will decompose
2x − 1
2
3
x + 6x + 5x − 12
into a sum of simpler fractions. First factor the denominator, then use the rules 1 through 4 to
write the form of a partial fractions decomposition:
2x − 1 2x − 1
=
x + 6x + 5x − 12 (x − 1)(x + 3)(x + 4)
2
3
A B C
= + + .
x − 1 x + 3 x + 4
Once we have this template, the rest is routine algebra. If the fractions on the right are added,
the numerator of the resulting quotient must equal 2x − 1, which is the numerator of the original
quotient. Therefore,
A(x + 3)(x + 4) + B(x − 1)(x + 4) + C(x − 1)(x + 3) = 2x − 1.

There are at least two ways we can find A, B, and C.
Method 1 Multiply the factors on the left and collect the coefficients of each power of x to
write

2
2
2
A(x + 7x + 12) + B(x + 3x − 4) + C(x + 2x − 3)
2
= (A + B + C)x + (7A + 3B + 2C)x + (12A − 4B − 3C) = 2x − 1.
Equate the coefficient of each power of x on the left to the coefficient of that power of x on the
right, obtaining a system of three linear equations in three unknowns:
2
A + B + C = 0 from the coefficients of x ,
7A + 3B + 2C = 2 from the coefficients of x,
and

12A − 4B − 3C =−1 from the constant term.
Solve these three equations obtaining A = 1/20, B = 7/4, and C =−9/5. Then

2x − 1 1 1 7 1 9 1
= + − .
2
3
x + 6x + 5x − 12 20 x − 1 4 x + 3 5 x + 4
Method 2 Begin with
A(x + 3)(x + 4) + B(x − 1)(x + 4) + C(x − 1)(x + 3) = 2x − 1,
and assign values of x that make it easy to determine A, B, and C. Put x = 1 to get 20A = 1,
so A = 1/20. Put x =−3 to get −4B =−7, so B = 7/4. And put x =−4 to get 5C =−9, so
C =−9/5. This yields the same result as method 1, but in this example, method 2 is probably
easier and quicker.

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October 14, 2010 14:14 THM/NEIL Page-119 27410_03_ch03_p77-120