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

May 18, 2015

We can rewrite this sum as

$\frac{{y}^{2} - 5}{{y}^{4} - 81} + \frac{4}{\left(- 1\right) \left({y}^{4} - 81\right)}$

We can find the lowest common denominator, which, then, is $\left(- 1\right) \left({y}^{4} - 81\right)$

Using the l.c.d.:

$\frac{\left(- 1\right) \left({y}^{2} - 5\right) + 4}{\left(- 1\right) \left({y}^{4} - 81\right)}$=$= \frac{- {y}^{2} + 9}{- {y}^{4} + 81}$

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:

$9 = {y}^{2}$
$y = \pm 3$, 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.

$\sqrt{{y}^{4}} = \sqrt{81}$

${y}^{2} = 9$

$y = \sqrt{9} = \pm 3$

So, here, as we are dealing with a polinomial of 4$t h$ degree, we can state that $- {y}^{4} + 81 = \left(x + 3\right) \left(x - 3\right) \left(x + 3\right) \left(x - 3\right)$, because we have ${y}^{\textcolor{g r e e n}{4}}$ and not only ${y}^{\textcolor{red}{2}}$ as in the previous factor.

$\frac{\cancel{x + 3} \cancel{x - 3}}{\cancel{x + 3} \cancel{x - 3} \left(x + 3\right) \left(x - 3\right)}$
Finalfinal answer, then: $\frac{1}{\left(x + 3\right) \left(x - 3\right)}$