# How do you express #(x^2)/((x^4) - (2x^2) - 8)# in partial fractions?

##### 1 Answer

#### Explanation:

If we let

#u/(u^2 - 2u - 8)#

The denominator can be factored as

#A/(u - 4) + B/(u + 2) = u/(u^2 - 2u - 8)#

#A(u + 2) + B(u - 4) = u#

#Au + 2A + Bu - 4B = u#

We can now write a system of equations:

#{(A + B = 1), (2A - 4B = 0):}#

Simplify the second equation to get:

#{(A + B = 1), (A - 2B = 0):}#

Substitute

#1 - B - 2B = 0#

#-3B = -1#

#B = 1/3#

This means that

Therefore, we can say

#u/(u^2 - 2u - 8) = 2/(3(u - 4)) + 1/(3(u + 2))#

Reversing our initial substitution:

#x^2/(x^4 - 2x^2 - 8) =2/(3(x^2 - 4)) + 1/(3(x^2 + 2))#

But we now see that

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

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

#Ax - 2A + Bx + 2B = 2/3#

Then we have a new system of equations:

#{(A + B = 0), (B - A= 1/3):}#

From the first equation we deduce

Hence:

#2/(3(x^2 - 4)) = 1/(6(x- 2)) - 1/(6(x+ 2)) #

Putting everything back together, we get:

#x^2/(x^4 - 2x^2 - 8) = 1/(6(x - 2)) - 1/(6(x+ 2)) + 1/(3(x^2 + 2))#

Hopefully this helps!