partial fraction decomposition?

Question

Hello everyone, I have partial fraction decomp. homework and i do not understand it, or even how to begin it, at all.

can anyone tell me how to do these few problems, and
i do not have to solve for constants.

1. ( x-2 ) / (x^2 + 4x + 3)

2. (6x+5) / (x+2) ^4

and now i have to write the partial fraction decomp of the rational expression...
1. (1) / (4x^2 - 9)

Answer 1

(x-2)/(x^2 + 4x + 3)
= (x-2)/{(x+1)(x+3)}
= A/(x+1) + B/(x+3)

Find A and B.

(x-2)/{(x+1)(x+3)} = N(x)/{(x-a) G(x)}
N(x) = x-2
G(x) = x+3
a = -1
A = N(a)/G(a) = N(-1)/G(-1)
A = (-1-2)/(-1+3) = -3/2

(x-2) / (x^2 + 4x + 3) = (-3/2)/(x+1) + B/(x+3)

Now we go the other way with G(x).

(x-2)/{(x+1)(x+3)} = N(x)/{(x-b) G(x)}
N(x) = x-2
G(x) = x+1
b = -3
N(x) / { (x-b) G(x) }
B = N(b)/G(b) = N(-3)/G(-3)
B = (-3-2)/(-3+1)
B = -5/(-2) = 5/2

Solution...

(x-2) / (x^2 + 4x + 3) = (5/2)/(x+3) - (3/2)/(x+1)

Answer 2

1. ( x-2 ) / (x^2 + 4x + 3)
= (x-2)/[(x+1)(x+3)]
= [A(x+1) + B(x+3)]/[(x+1)(x+3)]
Ax+A +Bx+3B = (A+B)x +A+3B = numerator
A+B = 1 = coefficient of x
A+3B=-2 = constant part of numerator
-2B =3 so B= - 1.5
A+B= 1 so A = 2.5

2. (6x+5) / (x+2) ^4
Do same procedure except now the fractions will be :
A/(x+2) +B/(x+2)^2 +C/ (x2)^3 + D/(x+2)^4

and now i have to write the partial fraction decomp of the rational expression...
1. (1) / (4x^2 - 9)
= 1/[(2x-3)(2x+3)]
=A(2x-3) + B(2x+3) = 2Ax -3A +2Bx +3B
= (2A+2B)x + 3B-3A
So 2A+2B=0 = coefficient of x of numerator and
3B-3A=1 = constant value of numerator
So A=-B
So 3B-3(-B) = 1
6B= 1 so B = 1/6 and A = -1/6

Answer 3

1.
( x-2 ) / (x^2 + 4x + 3)
= (x-2)/[(x+1)(x+3)]
= A/(x+1) + B/(x+3)
= [A(x+3) +B(x+1)]/(x^2 + 4x + 3)

Compare coefficients:
A+B = 1......(1)
3A+B = -2......(2)

(1)-(2): -2A = 3

A = -3/2, B = 5/2

( x-2 ) / (x^2 + 4x + 3) = (-3/2)/(x+1) + (5/2)/(x+3)

In the same way, you should know how to do problem 2.

Answer 4

See this site for detailed help:
http://jwbales.home.mindspring.com/precal/part7/part7.6.html

Hope it helps!