Question

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

Answer 1

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))`