Partial Fractions Method

In algebra the partial fraction method decomposition or partial fraction method expansion is used to reduce the degree of either the numerator or the denominator of a rational function. The outcome of a full partial fraction method expansion expresses that function as a sum of fractions. In symbols, one can use partial fraction method expansion to change a rational function in the form. (Source: wikipedia)

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General Notation of Partial Fractions method with Example:

General Steps to be followed in partial fractions:

Step 1: First we can divide the terms into partial fractions

Step 2: This is known as A and B

Step 3: We can plug any value for A

Step 4: We can get the value for A

Step 5: We can plug A value on the one of the equation

Step 6: Then we have to get the value of B

Step 7: These A and B are the solutions of this partial fractions.

General Notation and Example Problems on the Partial Fractions Method:

Problem 1:

If we wants to decompose (m)/(m+n) , then one can follow these steps:

Write as

(m)/(m+n) = A +(B)/(m+n)

Here A, B and n are known as constants.

Multiply both sides by m+n

m = A(m+n) +B

m = A.m + A.n +B

Here A and B must simultaneously solve:

m = A.m

0 = A.n + B

Because the coefficients of both m and 1 have to agree, giving rise to the two equations.

Therefore here A = -1 and B = -a

Finally, the decomposed form is:

(m)/(m+n) = 1+(-n)/(m+n)

These are the general notation of partial fractions method.

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Problem 2:

Simplify   5              into partial fractions.using partial fraction method
(x + 1)(x + 2)

Solution:

((5)/((x+1)(x+2)))

it can be written as

(A)/(x+1) + (B)/(x+2)

Taking L.C.M and simplifyning this we get,

5 = A(x+2) + B(x+1)

5 = Ax +2A + Bx +B

Plug x=-1 we get

5 = -A + 2A - B + B

here B get cancelled

Then

5 = A

Therefore A = 5

Again consider

5=Ax +2A + Bx +B

Plug x=-2 we get

5=-2A +2A +B(-2) +B

Here -2A and 2A get cancelled

5=-B

Therefore B=-5

Therefore the solution of these partial Equation is A=5and B=-5