Divide as indicated and simplify: (y^2 - 81) divided by the fraction (y^2 + 4y - 45)/(y^2 + 9).

Step-by-step solution

1. 1

   Turn the division into a multiplication first. Dividing by a fraction is multiplying by its reciprocal, so flip the second expression before touching anything else. Trying to cancel across a division bar is the most common mistake in this problem type:

   $\left(y^{2}-81\right)\div\frac{y^{2}+4y-45}{y^{2}+9}=\left(y^{2}-81\right)\cdot\frac{y^{2}+9}{y^{2}+4y-45}$

2. 2

   Factor every polynomial completely. Cancellation is only legal between factors, so nothing can be simplified until each piece is a product. The first is a difference of squares, and the trinomial factors from the pair that multiplies to $-45$ and adds to $4$, namely $9$ and $-5$:

   $y^{2}-81=(y-9)(y+9),\qquad y^{2}+4y-45=(y+9)(y-5)$

   Note that $y^{2}+9$ is a sum of squares and does not factor over the reals, so it stays as it is.

3. 3

   Assemble the single fraction. Put the factored numerator over the factored denominator:

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

4. 4

   Cancel the common factor and note the restrictions. The factor $(y+9)$ appears top and bottom, so it divides out:

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

   The cancellation is valid only where $y+9\neq 0$, and the remaining denominator forbids $y=5$, so the simplified form holds for $y\neq -9$ and $y\neq 5$.

5. 5

   Confirm the result with a test value. At $y=2$ the original expression is $(4-81)\div\dfrac{4+8-45}{4+9}=(-77)\cdot\dfrac{13}{-33}=\dfrac{1001}{33}$, and the simplified form gives $\dfrac{(2-9)(4+9)}{2-5}=\dfrac{-91}{-3}=\dfrac{91}{3}=\dfrac{1001}{33}$. The two agree, so the algebra is sound.