How do you add or subtract #(y^2-5)/(y^4-81) + 4/(81-y^4)#?

1 Answer
May 18, 2015

We can rewrite this sum as

#(y^2-5)/(y^4-81)+4/((-1)(y^4-81))#

We can find the lowest common denominator, which, then, is #(-1)(y^4-81)#

Using the l.c.d.:

#((-1)(y^2-5)+4)/((-1)(y^4-81))#=#=(-y^2+9)/(-y^4+81)#

That can be your final answer, but let's draw attention to the fact we have factorable functions there.

Let's find the roots of #-y^2+9# by equaling this to zero and, then, factoring it:

#9=y^2#
#y=+-3#, which means #y-3=0# and #y+3=0#. The two roots have been found.

Now, as for the denominator:

#y^4=81#

Let's just go slowly here and take the square root of both sides.

#sqrt(y^4)=sqrt(81)#

#y^2=9#

#y=sqrt(9)=+-3#

So, here, as we are dealing with a polinomial of 4#th# degree, we can state that #-y^4+81=(x+3)(x-3)(x+3)(x-3)#, because we have #y^color(green)4# and not only #y^color(red)2# as in the previous factor.

Therefore, your "final" answer can be rewritten as

#(cancel(x+3)cancel(x-3))/(cancel(x+3)cancel(x-3)(x+3)(x-3))#

Finalfinal answer, then: #1/((x+3)(x-3))#