## Friday, March 15, 2013

### Partial Fractions

Category: Algebra

"Published in Newark, California, USA"

Resolve into partial fractions for

Solution:

In this lesson, we need to learn this one because you will use this method often when you will study or take-up Integral Calculus. If you will integrate the algebraic fractions, then you must rewrite it into partial fractions first. If you know how to combine a fraction either by addition, subtraction, multiplication, or division, then you must know how to split a fraction into partial fractions. Anyway, let's consider the given equation above

The first thing that we have to do is to factor the numerator and denominator if you can. You must simplify a fraction into  lowest term always. That's a rule in Mathematics. As you notice that one of the factor in the denominator which is x2 + 3x + 3 cannot be factored, then we have to leave it as is. In this case, we have to split the above equation into partial fractions as follows

Multiply both sides of the equation by their Least Common Denominator (LCD) which is (2x + 3)(x2 + 3x + 3) as follows

Expand the right side of the equation and group according to their variables

In order to solve for the value of A, B, and C, we need to equate the both sides of the equation according to their variables.

For x2: 0 = A + 2B
A = - 2B                         (equation 1)

For x: 12 = 3A + 3B + 2C
12 = 3(- 2B) + 3B + 2C
12 = - 6B + 3B + 2C
12 = - 3B + 2C                (equation 2)

For x0: 21 = 3A + 3C
7 = A + C
7 = - 2B + C                 (equation 3)

Use equation 2 and equation 3 to solve for the value of B as follows

- 3B + 2C = 12     —————> - 3B + 2C = 12
-2(- 2B + C = 7)                             4B - 2C = - 14
————————
B = - 2

Substitute the value of B to equation 3, we have

- 2B + C = 7
- 2(- 2) + C = 7
4 + C = 7
C = 3

Substitute the value of B to equation 1, we have

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

Therefore,