Verified answer

There are different methods you can use. There is one way that doesn't involve solving any equations. This works as follows. We expand the function around all its singularities and keep only the singular terms. The sum of these expansions is then exactly the partial fractions expansion if the degree of the numerator is less than the degree of the denominator. I won't prove this here, but it's not difficult to prove.

So, in this case let's do the expansion about x = -1 first. Let's put x = -1 + t.

2(t-1)/[(t+3)t^2] =

2/t^2 (t-1)* 1/(t+3) =

Expand 1/(t+3) in powers of t:

1/(t+3) = 1/3 1/(1 + t/3) =

1/3 (1-t/3 + t^2/9 - ....)

So, the function becomes:

2/3 t^(-2) (t-1)* (1-t/3 + t^2/9 - ....) =

2/3 t^(-2) [-1 + 4/3 t + O(t^2)] =

-2/3 t^(-2) + 8/9 t^(-1) + non singular terms.

In terms of x, this is:

-2/3 1/(x+1)^2 + 8/9 1/(x+1)

Next we expand around the singularity at x = -4. If put

x = -4 + v, you get:

1/v 2(-4+v)/(-3+v)^2

We only need the zeroth order term in the expansion of 2(-4+v)/(-3+v)^2 as terms of order v yield non singular terms. So, all we have to do is to put v = 0 in there. This amounts to doing the computation as:

2X / (X+4) (X+1)^2 =

1/(x+4) 2x/(x+1)^2 =

1/(x+4) expansion of 2x/(x+1) around x = -4 to zeroth order.

expansion of 2x/(x+1)^2 around x = -4 to zeroth order = value this assumes at x = -4 = -8/(-3)^2 = -8/9

So, the expansion is 8/3 1/(x+4). The partial fraction expansion is thus:

--2/3 1/(x+1)^2 + 8/9 1/(x+1) - 8/9 1/(x+4)

a million) i'm assuming you propose (5x+12)/(x^2+5x+6). First, element the denominator. x^2 + 5x + 6. think of of two numbers that upload to 5 and multiply to 6. i will think of of two and 3. So x^2 + 5x + 6 = (x+2)(x+3) Now we'd desire to discover a and B such that A/(x+2) + B/(x+3) = (5x+12)/(x^2+5x+6). A/(x+2) + B/(x+3) = (5x+12)/(x^2+5x+6) Multiply all words by utilising (x^2+5x+6) to get rid of all the fractions. A(x+3) + B(x+2) = 5x + 12 Ax + 3A + Bx + 2B = 5x + 12 This finally leads to a pair of simultaneous linear equations. Ax + Bx = 5x A + B = 5 3A + 2B = 12 Use in spite of technique you desire to remedy . i will use substitution. A = 5 - B 3(5-B) + 2B = 12 15 - 3B + 2B = 12 -B = 12 - 15 = -3 B = 3 A = 5 - (3) = 2 consequently A/(x+2) + B/(x+3) = (5x+12)/(x^2+5x+6) 2/(x+2) + 3/(x+3) = (5x+12)/(x^2+5x+6) 2. i'm assuming you propose (5x-a million) / (x-a million)^2 The partial fractions might have denominators of (x-a million) and (x-a million)^2. If the denominator were (x-a million)^3, the partial fraction denominators could be (x-a million), (x-a million)^2 and (x-a million)^3. A/(x-a million) + B/(x-a million)^2 = (5x-a million) / (x-a million)^2 Multiply all words by utilising (x-a million)^2 to get rid of the fractions. A(x-a million) + B = 5x - a million Ax - 1A + B = 5x - a million Ax = 5x A = 5 -A + B = -a million -5 + B = -a million B = -a million + 5 = 4 5/(x-a million) + 4/(x-a million)^2 = (5x-a million) / (x-a million)^2

A/(x+4) + B/(x+1) + C/(x+1)^2

A(x+1)^2 +B(x+1)(x+4) + C(x+4)

A(x2+2x+1) + B(x^2+5x+4) + Cx + 4C

Compare the coeff

A+B = 0

2A + 5B + C = 2

A+4B+4C=0

Solve for A,B,C