Partial Fraction?

Question

How to partialise
1/(X+2)(X²+4)
the answer isn't 1/8(X+2) + 1/4(X²+4)

Answer 1

1/[(x + 2)(x^2 + 4)] has a linear factor and an irreducible quadratic factor. The partial fraction decomposition is this:

1/[(x + 2)(x^2 + 4)] = A/(x + 2) + (Bx + C)/(x^2 + 4)

Let's solve for A and B. Multiply both sides by (x + 2)(x^2 + 4), we get

1 = A(x^2 + 4) + (Bx + C)(x + 2)

Remember that this is true _for all x_. Let x = -2; then
1 = A( (-2)^2 + 4) + 0
1 = A(4 + 4)
1 = 8A, so A = (1/8).

Let x = 0: Then
1 = A(0 + 4) + C(0 + 2)
1 = 4A + 2C. But A = (1/8), so
1 = 4(1/8) + 2C
1 = 1/2 + 2C
1/2 = 2C, so C = 1.

Let x = 1: Then
1 = A(x^2 + 4) + (Bx + C)(x + 2)
1 = A(5) + (B + C)(3)
1 = (1/8)(5) + 3(B + 1)
1 = (5/8) + 3B + 3
8 = 5 + 24B + 24
8 = 29 + 24B
-21 = 24B
-21/24 = B, or
B = -7/8

A = 1/8, B = -7/8, C = 1, so our answer is

(1/8) (1/(x + 2)) + ((-7/8)x + 1) (1/(x^2 + 4))

Answer 2

A / (x+2) + (Bx+C) / (x^2 + 4) = 1/[ (X+2)(X²+4) ]
multiply by (X+2)(X²+4) to get
A(x^2+4) + (Bx+C) (x+2) = 1

Let x=-2 ====> 8A=1 ====> A = 1/8
Let x=0 ====> 4A + 2C = 1 ===> 4/8 + 2C = 1 ==> C= 1/4
(I get dif't answer B)
Let x = 2 ====> 8A + (2B+C)(4) = 1 ===> 1+8B +1 = 1 => B=-1/8

I get 1/8(x+2) + (-1 + 2x) / [ 8(x^2+4) ]

check by adding back togethr with LCD

Answer 3

Write it as
A / (x+2) + (Bx + C) / (x^2 + 4)
So we get
A(x^2 + 4) + (Bx + C) (x + 2) = 1
<=> x^2 (A + B) + x (2B + C) + 1 (4A + 2C) = 1
<=> A + B = 0, 2B + C = 0, 4A + 2C = 1.
So B = -A, C = -2B = 2A, 4A + 2C = 8A = 1.
So A = 1/8, B = -1/8, C = 1/4
and hence we have
1/[(x+2)(x^2+4)] = 1/[8(x+2)] + (2-x)/[8(x^2+4)].

Answer 4

go to google and type in
"partial fraction decomposition"

try this website or any other that google gives you:

http://www.mathematicshelpcentral.com/lecture_notes/precalculus_algebra_folder/partial_fraction_decomposition.htm