Can this problem be simplified further?
$\sqrt{(x+5)(x-7)}$
My initial thought was to set one side to zero and the square both sides but so I have been stuck. Its been way to long since I've had to do this and do not remember how to get rid of the radical sign.
On
This depends on the assumptions on the domain for $x$ and $y$. For real numbers, there is no simplification $(x+5)(x+7)=y^2$ in terms of $x$. However, should you be interested in finite fields, then over the field $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ we have $$ (x+5)(x+7)=(x+1)^2, $$ so its square root is simply $x+1$.
What you are asking about is the simplification of an expression $$\sqrt{(x+5)(x-7)}$$
There is no equation to solve here, so you can not pretend that it equals zero, and then solve for $x$. That means, you need to check whether $(x+5)(x-7)$ is perhaps a square.
But, note that $$(x + 5)(x - 7) = x^2 - 2x - 35 = (x^2 - 2x + 1) - 36 = (x - 1)^2 - 36$$ and hence, $(x+5)(x-7)$ is not a perfect square of the form $y^2$, so you're stuck with the radical remaining.
Hence, $$\sqrt{(x+5)(x-7)}$$ is as simplified as it gets.