plz Solve partial fracton= x / (x+2) (x(square) + 4)?

Question

its partial fraction
x
--------
(x+2)(x(sq) + 4)

Answer 1

First, we need to separate the denominator into linear factors and irreducible quadratic factors. Fortunately, this has already done; (x+2) is linear and (x^2 + 4) is obviously irreducible. (A quadratic is irreducible if it can be expressed as (x-a)^2 + b, where b is positive; here a = 0 and b = 4.)

So we want partial fractions of the form
A / (x+2) + (Bx + C) / (x^2 + 4)

Put all of this over a common denominator and we get
[A(x^2 + 4) + (Bx + C)(x+2)] / [(x+2)(x^2 + 4)]

So we want to have
A(x^2 + 4) + (Bx + C)(x + 2) = x, for all x
LHS = Ax^2 + 4A + Bx^2 + Cx + 2Bx + 2C
= (A+B) x^2 + (C+2B) x + (4A+2C)

Equating coefficients of x^2, x and 1 gives us
A + B = 0
C + 2B = 1
4A + 2C = 0

The first equation gives us B = -A and the third gives us C = -2A. Putting these into the second equation gives us -2A - 2A = 1, so A = -1/4, B = 1/4 and C = 1/2.

Thus the partial fraction is
(-1/4) / (x+2) + (x/4 + 1/2) / (x^2 + 4)
or
-1/(4x + 8) + (x+2)/(4x^2 + 16).

Answer 2

So you want to find the partial fraction decomposition of

x / [ (x+2) (x^2 + 4) ]

Note that the denominator contains a linear root, and an irreducible quadratic. The decomposition goes as follows.

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

Note that irreducible quadratics (i.e. ones that don't factor) always have a numerator of the form Bx + C, as opposed to the other factor, which has just an A.

Now, we multiply both sides by (x+2) (x^2 + 4) to eliminate all fractions. This leaves us with

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

Now, we expand that.

x = Ax^2 + 4A + Bx^2 + 2Bx + Cx + 2C

And now we combine like terms. It is important to ignore the left hand side.

x = Ax^2 + Bx^2 + 2Bx + Cx + 4A + 2C

And now, combine like terms.

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

I'm going to change the left hand side to make something obvious.

0x^2 + x + 0 = (A+B)x^2 + (2B + C)x + (4A + 2C)

What you have to do at this point is **pair up**, on the left hand side and the right hand side, the coefficients of each of x^2, x, and the constant.

A + B = 0
2B + C = 1
4A + 2C = 0

(Note: the coefficient of x on the left hand side is 1, but we usually don't write 1x, therefore it's 1).

Now, we solve that system of equations for A, B, and C.
I'll leave you to do that, but the solution we get is

A = -1/4, B = 1/4, C = 1/2

So the partial fraction decomposition is equal to those values plugged in. That is

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

Turns into

x / [ (x+2) (x^2 + 4) ] = (-1/4)/(x+2) + ([1/4]x + [1/2])/(x^2 + 4)

Answer 3

If you are trying to join these two then the answer would be

x (X^2 +4)
____________
X+2