The Four Cases of Partial Fractions Resolution

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-a_n)

 Where,

 a₁≠a₂≠a₃≠⋯.a_n

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

Where, the coefficient A₁,A₂,A₃,….A_n  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+ ….+A_n/(x-a_n)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,….A_n 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+……+A_nx+B_n/(ax²+bx+c)ⁿ

Where A₁,B₁,A₂,B₂,………,A_n,B_n  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)

Frequently Asked Question-FAQs