# Express 1/(x^2 – 4)(x+3) as a sum of fractions.

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### 2 Answers

We have to express 1/(x^2 – 4)(x+3) as a sum of fractions.

Now 1/(x^2 – 4)(x+3)

=> 1/ (x – 2)(x + 2)(x + 3)

=> A / (x – 2) + B/ (x + 2) + C / (x + 3)

=> [A(x^2 + 5x + 6) + B ( x^2 + x – 6) + C (x^2 – 4)]/ (x – 2)(x + 2)(x + 3)

Now equating the coefficients of x^2, x and the numeric terms we get three equations:

A + B + C = 0 …(1)

5A + B = 0 …(2)

6A – 6B – 4C = 1 …(3)

From (2)

5A + B = 0

=> **B = -5A**

Substitute this in (1) and (3)

A + B + C = 0

=> -4A + C = 0

=> **C = 4A**

6A – 6B – 4C = 1

=> 6A + 30A – 4C = 1

substitute C = 4A

=> 36A – 4*(4A) = 1

=> 20A = 1

=> **A = 1/20**

**C = 4A = 1/5**

**B = -5/20 = -1/4**

**Therefore 1/(x^2 – 4)(x+3) = 1 / 20*(x – 2) - 1/ 4*(x + 2) + 1 / 5*(x + 3)**

1/(x^2 - 4) ( x+3)

We need to rewrite using partial fractions.

Let us simplify.

==> Let E = 1/(x-2)(x+2)(x+3)

==> we will assume that:

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

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

==> A( x^2 + 5x + 6) + B(x^2 + x - 6) + C(x^2 -4) = 1

==> (A+B+C)x^2 + (5A+ B)x + (6A-6B -4C) = 1

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

==> 5A+B = 0...............(2)

==> 6A-6B-4C = 1.............(3)

Now we will solve the system.

==> From (2) we know that B = -5A

Now we will substitute into (1) and (3).

==> A - 5A + C = 0

==> -4a + C = 0................(4)

==> 6A-6(-5A) - 4C = 1

==> 6A + 30A - 4C = 1

==> 36A - 4C = 1............(5)

We will multiply (4) by 9 and add to (5).

==> -36A + 9C = 0

==> 36A - 4C = 1

==> 5C = 1

**==> C = 1/5**

==> -4A + C = 0

==> -4A + 1/5 = 0

==> A = -1/5 / -4 = 1/20

**==> A = 1/20**

==> B = -5A = -5(1/20) = -5/20 = -1/4

**==> B = -1/4**

**==> 1/(x^2-4)(x+3) = 1/20(x-2) - 1/4(x+2) +1/5(x+3)**