**Partial fraction decomposition examples with answers :**

Here we are going to see some example problems on partial fractions.

A given proper fraction can be expressed as the sum of other simple fractions corresponding to the factors of the denominator of the given proper fraction. This process is called splitting into Partial Fractions.

Proper fraction :

A proper fraction is one in which the degree of the numerator is less than the degree of the denominator.

When the factors of the denominator of the given fraction are all linear factors none of which is repeated. We write the partial fraction as follows.

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

here the denominator is in the form linear factors and no factor is repeated. So we can write the partial-fraction as

**(x +3)/(x + 1) (x - 2) = [A/(x + 1)] + [B/(x - 2)]**

where A and B are constants.

If a linear factor (a x + b) occurs n times as a factor of the denominator of the given fraction, then we can write the partial-fraction as

**(x +3)/(x-2)³ = [A/(x - 2)]+[B/(x - 2)²]+[C/(x - 2)³]**

where A, B and C are constants.

If a quadratic equation a x² + b x + c which is not favorable into linear factors occurs only once as factor of the denominator of the given fraction, then we can write the partial fraction as

**(x +3)/(x-1) (x****²+2x-2)=[A/(x-1)]+[(Bx+c)/****(x****²+2x-2)]**

If we have only one quadratic equation alone in the denominator we can write it as simply (Ax + B)/(the given quadratic equation)

**Example 1 :**

Resolve into partial fractions

**Solution :**

1/(x - 1) (x + 1) = A/(x - 1) + B/(x + 1)

1/(x - 1) (x + 1) = A(x + 1) + B(x - 1)/(x - 1)(x + 1)

Since we have same denominators on both side, we can equate the numerators.

1 = A(x + 1) + B(x - 1) ------(1)

The constants A and B can also be found by successively giving suitable values for x.

To find A, put x = 1 in (1)

To find B, put x = -1 in (1)

1 = A(1 + 1) + B(1 - 1) 1 = 2A + B (0) 1 = 2A + 0 2A = 1 ===> A = 1/2 |
1 = A(-1 + 1) + B(-1 - 1) 1 = A(0) + B (-2) 1 = 0 - 2B -2B = 1 ===> B = -1/2 |

Let us look into the next example on "Partial fraction decomposition examples with answers".

**Example 2 :**

Resolve into partial fractions

**Solution :**

7x - 1/(x^{2}-5x + 6) = 7x - 1/(x - 3) (x - 2)

7x - 1/(x - 3) (x - 2) = A/(x - 3) + B/(x - 2)

7x - 1/(x - 3) (x - 2) = A(x - 2) + B(x - 3)/(x - 3)(x - 2)

Since we have same denominators on both side, we can equate the numerators.

7x - 1 = A(x - 2) + B(x - 3) ------(1)

The constants A and B can also be found by successively giving suitable values for x.

To find A, put x = 3 in (1)

To find B, put x = 2 in (1)

7(3) - 1 = A(3-2)+B(3-3) 21 - 1 = 1A + B (0) 20 = A + 0 A = 20 |
7(2) - 1 = A(2-2)+B(2-3) 14 - 1 = A(0) + B (-1) 13 = 0 - B B = -13 |

Let us look into the next example on "Partial fraction decomposition examples with answers".

**Example 3 :**

Resolve into partial fractions

**Solution :**

(x^{2}+ x + 1)/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3)

Taking L.C.M

= A(x-2)(x-3) + B(x-1)(x-3) + C(x-2)(x-1)/(x-1)(x-2)(x-3)

Since we have same denominators on both side, we can equate the numerators.

x^{2}+ x + 1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-2)(x-1) ----(1)

The constants A, B and C can also be found by successively giving suitable values for x.

To find A, put x = 1 in (1)

To find B, put x = 2 in (1)

To find C, put x = 3 in (1)

If we put x = 1, both B and C will become zero. |
1 3 = A(-1)(-2) 2A = 3 ==> A = 3/2 | |

If we put x = 2, both A and C will become zero. |
2 4 + 2 +1 = B(1)(-1) 7= -B ==> B = -7 | |

If we put x = 3, both C and B will become zero. |
3 9 + 3 + 1 = C(1)(2) 13 = 2C ==> C = 13/2 |

**Let us look into the next example on "**Partial fraction decomposition examples with answers**".**

**Example 4 :**

Resolve into partial fractions

**Solution :**

1/(x-1)(x+2)^{2} = A/(x-1) + B/(x+2) + C/(x+2)^{2}

Taking L.C.M

1/(x-1)(x+2)^{2 }= A(x+2)^{2} + B(x-1)(x+2) + C(x-1)/(x-1)(x+2)^{2}

Since we have same denominators on both side, we can equate the numerators.

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

By equating the coefficients of x^{2}, x and constant terms respectively, we get

A + B = 0 -------(2)

4A + B + C = 0 -------(3)

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

By adding the equations (3) and (4) we get,

00000000000004A + B + C = 000000000000000000000

00000000000004A - 2B - C = 10000000000000000000

000000000000-------------------0000000000000000000

0000000000008A - B = 1------(5)00000000000000000

(5) + (2)

A + B + 8A - B = 0 + 1

9A = 1 ==> A = 1/9

Substituting the value of A in (2), we get 1/9 + B = 0 B = -1/9 |
Substituting the values of A and B in (3), we get 4(1/9) + (-1/9) + C = 0 4/9 - 1/9 + C = 0 (4 - 1 + 9C)/9 = 0 3 + 9C = 0 9C = -3 C = -3/9 ==> -1/3 |

**Example 5 :**

Resolve into partial fractions

**Solution :**

(x - 2)/(x+2)(x-1)^{2} = A/(x + 2) + B/(x - 1) + C/(x - 1)^{2}

Taking L.C.M

(x-2)/(x+2)(x-1)^{2} = A(x-1)^{2}+B(x-1)(x+2)+C(x+2)/(x+2)(x-1)^{2}

Since we have same denominators on both side, we can equate the numerators.

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

By equating the coefficients of x^{2}, x and constant terms respectively, we get

A + B = 0 -------(2)

-2A + B + C = 1 -------(3)

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

2 x (3) - (4)

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

-4A - A + 2B + 2B + 2C - 2C = 2 + 2

-5A + 4B = 4 -------(5)

5 x (2) + (5)

5A + 5B + (-5A + 4B) = 0 + 4

5A + 5B -5A + 4B = 4

9B = 4 ==> B = 4/9

Substituting the value of B in (2), we get A + 4/9 = 0 A = -4/9 |
Substituting the values of A and B in (3), we get -2(-4/9) + 4/9 + C = 1 8/9 + 4/9 + C = 1 (8 + 4 + 9C)/9 = 1 12 + 9C = 9 9C = -3 C = -1/3 |

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