Home | Math | The Four Cases of Partial Fractions Resolution

# The Four Cases of Partial Fractions Resolution

July 13, 2022 written by Azhar Ejaz

To express a single rational fraction as a sum of two or more two rational fractions is called a partial fraction. There are Four Cases of Partial Fractions Resolution

## Resolution of p(x)/Q(X) into partial fractions when Q(X) has only non–repeated linear factors.

The polynomial Q(x) may be written as:

Q(x) =(x-a1)(x-a2)(x-a3)………..(x-an)

Where,

a1≠a2≠a3≠⋯.an

∴P(x)/Q(X) =A1/(x-a1)+A2/(x-a2)+…+An/(x-an) is an identity.

Where, the coefficient A1,A2,A3,….An  are numbers to be found .

The method is explained by the following

Example:

Resolve, 7x+25/(x+3)(x+4) into partial fractions.

Solution:

Suppose

7x+25/(x+3)(x+4) = A/(x+3)+B/(x+4)……(1)

Multiplying both sides by (x+3)(x+4) equation(1) we get,

7x+25 =A(x+4)+B(x+3)…..(2)

As two sides identity are equal for all values of x,

Let us put x+3=0 ,

X=-3 in equation (2)

7(-3)+25=A(-3+4)+B(-3+3)

-21+25=A(1)+B(0)

4=A+0

4=A

Putting x+4=0 ,x=-4 in equation(2)

7(-4)+25=A(-4+4)+B(-4+3)

-28+25=A(0)+B(-1)

-3=0-B

-3=-B

3=B

Put the value of constant A and B in equation (1)

7x+25/(x+3)(x+4)=4/(x+3)+3/(x+4)

## When Q(x) has repeated linear factors:

P(x)/Q(X)=A1/(x-a1)1+A2/(x-a2)2+ ….+An/(x-an)n

if the polynomial has a factor (x-a)n, n≥2, and n is a positive integer, then P(X)/Q(X) may be written as the following  identity :

Where the coefficients A1, A2, A3,….An are numbers to be found. The method is explained by the following

Example:

Resolve,   x2+x-1/(x+2)3 into partial fractions.

Solution:

Suppose

x2+x-1/(x+2)3 =A/(x+2)+B/(x+2)2+c/(x+2)3……(1)

Multiplying both side by(x+2)3   equation (1)

X2+x-1=A(x+2)2 +B(x+2)+C…….(2)

Put x+2=0,

X=-2 in equation in (2)

(-2)2-2-1=A(-2+2)2+B(-2+2)+C

4-2-1=A(0)+B(0)+C

1=0+0+C

1=C

From equation (2)

X2+x-1=A(x2+4x+4)+B(x+2)+C……(3)

Equating the coefficient of x2 and x in (3) we get,

1=A…….(a)

1=4A+B……(b)

Put the value of (A) in equation b

1=4(1)+B

1=4+B

1-4=B

-3=B

Put the value A,B and C in equation (1)

x2+x-1/(x+2)3=1/(x+2)-3/(x+2)2+1/(x+2)3

## When Q(x) contains a non-repeated irreducible quadratic factor.

A quadratic factor is irreducible if it cannot be written as the product of two linear factors with real coefficients. For example, x2+x+1 and x2+3 are irreducible quadratic factors.

If the polynomial Q(x) contains a non-repeated irreducible quadratic factor then P(x)/Q(x) may be written as the identity having partial fractions of the form   Ax + B/ax2+bx+c  where A and B are the numbers to be found. the method is explained by the following

Example:

Resolve 3x-11/(x2+1)(x+3) into partial fractions.

Solution:

Suppose:

3x-11/(x2+1)(x+3)=Ax+B/x2+1 +C/(x+3) …..(1)

Multiplying both side (x2+1)(x+3) by equation(1),we get

3x-11= (Ax +B) (x+3) +C(x2+1)………(2)

Put x+3=0

X=-3

In equation (2)

3(-3)-11 = (Ax+B)(-3+3)+C((-3)2+1)

-9-11=(Ax+B)(0)+C(9+1)

-20=10C

C=-20/10

C=-2

From equation (2)

3x-11=Ax2+3Ax+Bx+3B+Cx2+C……..(3)

Comparing the coefficient of x2 and x in equation ( 3)

A+C=0…..(a)

3A+B=3……(b)

Put the value C =-2 in equation (a)

A-2=0

A=2

Put the value A=2 in equation (b)

3(2)+B=3

6+B=3

B=3-6

B=-3

Put the value A,B,C in equation (1)

3x-11/(x2+1)(x+3)=2x-3/x2+1 -2/x+3

## When Q(x) has repeated irreducible quadratic factors:

if the polynomial Q(x) contains a repeated irreducible quadratic factor.

(ax2+bx+C)n, n≥2 and n is a  positive integer, than  P(x)/Q(x) may be written as the following identity:

P(x)/Q(x)=A1x+B1/(ax2+bx+c)+A2x+B2/(ax2+bx+c)2+……+Anx+Bn/(ax2+bx+c)n

Where A1,B1,A2,B2,………,An,Bn  are numbers to be found  .the method is explained through the following

Example:

Resolve 4x2/(x2+1)2(x-1) into partial fractions:

Solution:

Suppose

4x2/(x2+1)2(x-1)=Ax+B/(x2+1)+Cx+D/(x2+1)2+E/(x-1)…(1)

Multiplying both side(x2+1)2(x-1) by equation (1) we get,

4×2=(Ax+B)(x2+1)(x-1)+(Cx+D)(x-1)+E(x2+1)2…..(2)

4x2=(A+E)x4+(-A+B)x3+(A-B+C+2E)x2+(-A+B-C+D)x+(-B-D+E)………….(3)

Putting x-1=0

x=1 in (2) we get

4=E(1+1)2

4=E(2)2

4=4E

E=4/4

E=1

Equating the coefficient of x4,x3,x2,x,in (3) we get

0=A+E….(a)

0=-A+B….(b)

4=A-B+C+2E…..(c)

D=A-B+C……..(d)

Put the value of E=1 in equation (a)

0=A+1

A=-1

Put the value of A =-1 in equation (b)

O=-(-1)+B

B=-1

Put the value of A,B,E in equation (c)

4=-1-(-1)+C+2(1)

4=-1+1+C+2

C=2

Put the value A,B,C in equation  (d)

D=-1+1+2

D=2

Put the value A,B,C,D,E

4x2/(x2+1)2(x-1)=-x-1/(x2+1)+2x+2/(x2+1)2+1/(x-1)