Divide (9 − a²)/(a² − 8a + 15) by (2a² − 10a)/(a² − 10a + 25) and simplify.

Step-by-step solution

1. 1

   Factor every quadratic, watching the reversed difference of squares. The numerator $9-a^{2}$ is $a^{2}-9$ with the sign flipped, so it is essential to record the minus:

   $9-a^{2}=-(a-3)(a+3)$

   $a^{2}-8a+15=(a-3)(a-5),\qquad 2a^{2}-10a=2a(a-5),\qquad a^{2}-10a+25=(a-5)^{2}$

2. 2

   Turn the division into multiplication by the reciprocal.

   $\frac{-(a-3)(a+3)}{(a-3)(a-5)}\cdot\frac{(a-5)^{2}}{2a(a-5)}$

3. 3

   Cancel the first pair of factors. The $(a-3)$ cancels, leaving

   $\frac{-(a+3)}{a-5}\cdot\frac{(a-5)^{2}}{2a(a-5)}$

4. 4

   Cancel the (a − 5) factors. There are two on top and two on the bottom, so they cancel completely:

   $\frac{-(a+3)}{2a}$

   $\boxed{-\dfrac{a+3}{2a}}\qquad(a\neq 0,3,5)$

5. 5

   Check with a test value. At $a=1$: the original is $\dfrac{8}{8}\div\dfrac{-8}{16}=1\div\left(-\tfrac12\right)=-2$, and the answer gives $-\tfrac{4}{2}=-2$ ✓. Dropping the minus from $9-a^{2}$ would have produced $+2$, so the substitution check catches that error immediately.