**Partial Fractions Examples :**

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

**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 |

**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 |

**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 |

**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 |

After having gone through the stuff given above, we hope that the students would have understood, how to decompose a rational expression into partial fractions.

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