# Decompose into partial fractions : (8x + 14) / (x + 1)(x + 5)

### 3 Answers | Add Yours

We have to write (8x + 14) / (x + 1)(x + 5) as partial fractions

(8x + 14) / (x + 1)(x + 5) = A / (x +1) + B / (x + 5)

=> (8x + 14) / (x + 1)(x + 5) = A(x + 5) + B(x + 1) / (x +1)(x + 5)

=> 8x + 14 = Ax + 5A + Bx + B

=> A + B = 8 and 5A + B = 14

=> B = 8 - A

Substituting in 5A + B = 14

=> 5A + 8 - A = 14

=> 4A = 6

=> A = 6/4 = 3/2

B = 8 - 3/2 = 13/2

Therefore we have (8x + 14) / (x + 1)(x + 5) =

**[3 / 2(x +1)] + [13 / 2(x + 5)]**

Given the fraction (8x+14)/ (x+1)(x+5)

We need to rewrite into the form A/(x+1) + B/(x+5)

==> (8x+14)/ (x+1)(x+5) = A/(x+1) + B/(x+5)

We will rewrite with the common denominator.

==> (8x+14)/(x+1)(x+5) = A(x+5)+ B(x+1)]/ (x+1)(x+5)

Now we will multiply by (x+1)(x+5) so the denominator cancels.

==> 8x + 14 = A(x+5) + B(x+1)

==> 8x + 14 = Ax + Bx + 5A + B

==> 8x + 14 = (A+B)x + (5A+B)

Now we will compare similar terms.

==> A+B = 8............(1)

==> 5A+B = 14 ........(2)

Now we will solve the system.

We will subtract (1) from (2).

==> 4A = 6

==> A = 6/4 = 3/2

==> A = 3/2

But A+B= 8

==> B = 8- A = 8 - 3/2 = 13/2

==> B = 13/2

Now we will substitute:

==> A/(x+1) + B(x+5) = (3/2)/(x+1) + (13/2)/(x+5)

= 3/2(x+2) + 13/2(x+5)

**==> (8x+14)/(x+1)(x+5) = 3/2(x+1) + 13/2(x+5)**

To decompose into partial fractions: (8x + 14) / (x + 1)(x + 5).

Let (8x + 14) / (x + 1)(x + 5) = A/(x+1) + B/(x+5).

We multiply both sides by (x+1)(x+5):

8x+14 = A(x+5)+B(x+1)....(1). Since this is an identity, we put x= -1 in the identity (1):

8(-1)+14 = A(-1+5)+B(-1+1).

-8+14 = 4A. Or A = 6/4 = 3/2.

A= 3/2.

Now we put x = -5 in (1):

8(-5)+14 = A(-5+5)+B(-5+1)

-40+14 = -4B.

-26 = -4B, or B = -26/-4. = 13/2

**Therefore (8x+14)/(x+1)(x+5) = 3/{2(x+1)}+13/{2(x+5)}.**