# How do you express #3/ (x^2-6x+8)# in partial fractions?

##### 1 Answer

Jul 26, 2016

#### Explanation:

#3/(x^2-6x+8) = 3/((x-4)(x-2)) = A/(x-4)+B/(x-2)#

Then, using Heaviside's cover up method, we find:

#A=3/(4-2) = 3/2#

#B=3/(2-4) = -3/2#

So:

#3/(x^2-6x+8) = 3/(2(x-4))-3/(2(x-2))#

Spelled out a little slower:

Given:

#3/((x-4)(x-2)) = A/(x-4)+B/(x-2)#

Multiply both sides by

#3/(x-2) = A+B(x-4)/(x-2)#

Then let

#3/(4-2) = A + 0#

hence

Similarly find

Note that this is actually division by