RESOLVE INTO PARTIAL FRACTIONS

Express the following as a sum of partial fractions.

Problem 1 :

Solution:

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

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

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

6 - 1 = A(0) + B(5)

5 = 5B

B = 1

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

-4 - 1 = A(-5) + B(0)

-5 = -5A

A = 1

Problem 2 :

Solution:

2x + 5 = A(x + 1) + B(x - 2) ---> (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 = 2 in (1)

2(-1) + 5 = A(-1 + 1) + B(-1 - 2)

-2 + 5 = A(0) + B(-3)

3 = -3B

B = -1

2(2) + 5 = A(2 + 1) + B(2 - 2)

4 + 5 = A(3) + B(0)

9 = 3A

A= 3

Problem 3 :

Solution:

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

3 = A(2x - 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 = 1/2 in (1)

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

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

3 = A(0) + B(-1/2)

3 = -1/2B

B = -6

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

3 = A(1) + B(0)

A = 3

Problem 4 :

Solution:

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

1 = A(x - 2) + B(x + 4) ---> (1)

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

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

To find B, put x = -4 in (1)

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

1 = A(0) + B(6)

1 = 6B

B =1/6

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

1 = A(-6) + B(0)

1 = -6A

A = -1/6

Problem 5 :

Solution:

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

2x + 3 = A(x + 5) + B(x - 2) ---> (1)

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

To find A, put x = -5 in (1)

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

2(-5) + 3 = A(-5 + 5) + B(-5 - 2)

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

-7 = -7B

B = 1

2(2) + 3 = A(2 + 5) + B(2 - 2)

4 + 3 = A(7) + B(0)

7 = 7A

A = 1

Problem 6 :

Solution:

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

2x + 5 = A(x + 3) + B(x + 2) ---> (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)

2(-3) + 5 = A(-3 + 3) + B(-3 + 2)

-6 + 5 = A(0) + B(-1)

-1 = -B

B = 1

2(-2) + 5 = A(-2 + 3) + B(-2 + 2)

-4 + 5 = A(1) + B(0)

A = 1

RESOLVE INTO PARTIAL FRACTIONS

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