PARTIAL FRACTIONS WITH QUADRATIC FACTORS

Problem 2 :

Solution:

x² - x + 2 = A(x² + 3) + B(x + 1)

x² - x + 2 = Ax² + 3A + Bx + B

Equating the coefficient of x², x and constant terms, we get

A = 1

B = -1

Problem 3 :

Solution:

3x + 7 = A(x² + 1) + (Bx + C)(x + 3)

Put x = -3,

3(-3) + 7 = A(10) + 0

-9 + 7 = 10A

10A = -2

A = -1/5

Equating the coefficient of x², x and constant terms, we get

3x + 7 = Ax² + A + Bx² + 3Bx + Cx + 3C

A + B = 0 ---> (1)

3B + C = 3 ---> (2)

A + 3C = 7 ---> (3)

By applying value A in (1),

-1/5 + B = 0

B = 1/5

By applying value B in (2),

3(1/5) + C = 3

3/5 + C = 3

C = 3 - 3/5

C = 12/5

Problem 4 :

Solution:

1 = A(x² + 1) + (Bx + C)(x + 1)

Put x = -1,

1 = A(2) + 0

1 = 2A

A = 1/2

Equating the coefficient of x², x and constant terms, we get

1 = Ax² + A + Bx² + Bx + Cx + C

A + B = 0 ---> (1)

B + C = 0 ---> (2)

A + C = 1 ---> (3)

By applying value A in (3),

1/2 + C = 1

C = 1/2

By applying value B in (2),

B + 1/2 = 0

B = -1/2

Problem 5 :

Solution:

3x + 7 = A(x + 2)(x² + x + 1) + B(x - 2)(x² + x + 1) + (Cx + D)(x - 2)(x + 2)

Put x = 2,

3(2) + 7 = A(4)(7)

13 = 28A

A = 13/28

Put x = -2,

3(-2) + 7 = 0 + B(-4)(4 - 2 + 1) + 0

-6 + 7 = -4B(3)

1 = -12B

B = -1/12

3x + 7 = A(x + 2)(x2 + x + 1) + B(x - 2)(x2 + x + 1) + (Cx + D)(x - 2)(x + 2)

3x + 7 = A(x³ + x² + x + 2x² + 2x + 2) + B(x³ + x² + x - 2x² - 2x - 2) + (Cx + D)(x² - 4)

Equating the coefficient of x³, x², x and constant terms, we get

A + B + C = 0 ---> (1)

3A - B + D = 0 ---> (2)

3A - B - 4C = 3 ---> (3)

2A - 2B - 4D = 7 ---> (4)

By applying values A and B in (1),

By applying values A and B in (2),