Find the Symmetry f(x)=2/(x^2+2x-15)

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle -15}$ and whose sum is ${\displaystyle 2}$.

Write the factored form using these integers.

Find ${\displaystyle f(-x)}$ by substituting ${\displaystyle -x}$ for all occurrence of ${\displaystyle x}$ in ${\displaystyle f\left(x\right)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -x}$.

Rewrite ${\displaystyle -3}$ as ${\displaystyle -1\left(3\right)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -\left(x\right)-1\left(3\right)}$.

Rewrite ${\displaystyle -(x+3)}$ as ${\displaystyle -1(x+3)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -x}$.

Rewrite ${\displaystyle 5}$ as ${\displaystyle -1(-5)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -\left(x\right)-1(-5)}$.

Simplify the expression.

Check if ${\displaystyle f(-x)=f\left(x\right)}$.

Since ${\displaystyle \frac{2}{(x+3)(x-5)}}$${\displaystyle \ne }$${\displaystyle \frac{2}{(x-3)(x+5)}}$, the function is not even.

Multiply ${\displaystyle -1}$ by ${\displaystyle \frac{2}{(x-3)(x+5)}}$.

Since ${\displaystyle \frac{2}{(x+3)(x-5)}}$${\displaystyle \ne }$${\displaystyle -\frac{2}{(x-3)(x+5)}}$, the function is not odd.