# 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-a_{1})(x-a_{2})(x-a_{3})………..(x-a_{n})

Where,

a_{1}≠a_{2}≠a_{3}≠⋯.a_{n}

∴P(x)/Q(X) =A_{1}/(x-a_{1})+A_{2}/(x-a_{2})+…+A_{n}/(x-a_{n}) is an identity.

Where, the coefficient A_{1},A_{2},A_{3},….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)=A_{1}/(x-a_{1})^{1}+A_{2}/(x-a_{2})^{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 A_{1}, A_{2}, A_{3},….A_{n} are numbers to be found. The method is explained by the following

**Example**:

Resolve, x^{2}+x-1/(x+2)^{3} into partial fractions.

**Solution:**

Suppose

x^{2}+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)

X^{2}+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)

X^{2}+x-1=A(x^{2}+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)

**x ^{2}+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, x^{2}+x+1 and x^{2}+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/(x^{2}+1)(x+3) into partial fractions.

Solution:

Suppose:

3x-11/(x^{2}+1)(x+3)=Ax+B/x^{2}+1 +C/(x+3) …..(1)

Multiplying both side (x^{2}+1)(x+3) by equation(1),we get

3x-11= (Ax +B) (x+3) +C(x^{2}+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=Ax^{2}+3Ax+Bx+3B+Cx^{2}+C……..(3)

Comparing the coefficient of x^{2} 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/(x ^{2}+1)(x+3)=2x-3/x^{2}+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)=A _{1}x+B1/(ax^{2}+bx+c)+A_{2}x+B^{2}/(ax^{2}+bx+c)^{2}+……+A_{n}x+B_{n}/(ax^{2}+bx+c)^{n}**

Where A_{1},B_{1},A_{2},B_{2},………,A_{n},B_{n} are numbers to be found .the method is explained through the following

**Example:**

Resolve 4x^{2}/(x^{2}+1)^{2}(x-1) into partial fractions:

**Solution:**

Suppose

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

Multiplying both side(x^{2}+1)^{2}(x-1) by equation (1) we get,

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

4x^{2}=(A+E)x^{4}+(-A+B)x^{3}+(A-B+C+2E)x^{2}+(-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 x^{4},x^{3},x^{2},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

**4x ^{2}/(x^{2}+1)^{2}(x-1)=-x-1/(x^{2}+1)+2x+2/(x^{2}+1)^{2}+1/(x-1)**

## Frequently Asked Question-FAQs

### What is partial fraction math?

Partial fractions are the fractions that are used in the decomposition of a rational expression. So, when an algebraic expression is split into a sum of two or more rational expressions, each part is referred to as a partial fraction.

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