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Partial Fraction Calculator is a very useful tool. It basically uses decomposition method to evaluate the

unknown variable in the equation. The equation generally in case of this is less typical, but this method

involves steps which makes it very easy to evaluate such difficult problems.

This tool is having unique method which is different from derivative in the sense of its unique methodology.

And this methodology is very useful for finding the unknown variable from typical equations.

**Example 1:- **

Find the partial fraction of the below mentioned function:-

unknown variable in the equation. The equation generally in case of this is less typical, but this method

involves steps which makes it very easy to evaluate such difficult problems.

This tool is having unique method which is different from derivative in the sense of its unique methodology.

And this methodology is very useful for finding the unknown variable from typical equations.

Find the partial fraction of the below mentioned function:-

(x-1)(x+2)

**Solution 1:- **

(x-1)(x-2) (x-1) (x+2)

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

Making x = -2

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

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

-1 = B (-3)

Divide both sides by 3, we get

B = 1/3.

Now making x = +1,

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

5 = 3 A + B (0)

So 3A = 5

Divide both sides by 3, we get

A = 5/3.

So,

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

Making x = -2

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

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

-1 = B (-3)

Divide both sides by 3, we get

B = 1/3.

Now making x = +1,

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

5 = 3 A + B (0)

So 3A = 5

Divide both sides by 3, we get

A = 5/3.

So,

(x-1)(x-2) (x-1) (x+2)

**Example 2:- **

Find the partial fraction of the below mentioned function:-

Find the partial fraction of the below mentioned function:-

(x+1)(x-2)

**Solution 1:- **

(x-1)(x-2) (x+1) (x-2)

2X+1 = A (x-2) + B (x+1)

Making x = 2

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

5 = A (0) + B (3)

5 = B (3)

Divide both sides by 3, we get

B = 5/3.

Now making x = -1,

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

-1 = -3 A + B (0)

So A = 1/3

So,

2X+1 = A (x-2) + B (x+1)

Making x = 2

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

5 = A (0) + B (3)

5 = B (3)

Divide both sides by 3, we get

B = 5/3.

Now making x = -1,

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

-1 = -3 A + B (0)

So A = 1/3

So,

(x-1)(x-2) (x+1) (x-2)